Alex Chiang's IB Biology Blog

Topic 9 [HL]: Plants

noreply@blogger.com (Alex Chiang) — Sun, 04 May 2014 08:43:00 +0000

Task One

State the function of photosystem II, photosystem I and the calvin cycle in the production of glucose and sucrose.

Photosynthesis is a two step process with light dependent and independent reactions. Light dependent reactions convert light energy to chemical energy and light independent reactions use the chemical energy to make organic molecules.

Light dependent reactions occur on thylakoid membrane and may occur via cyclic or non-cyclic processes; both processes include light exciting chlorophyll leading to release of electrons that pass through an electron transport chain, creating ATP (photophosphorylation)

Non-Cyclic: Chlorophyll in photosystems 1 and 2 absorbs light, leading to photoactivation: a release of high energy electrons. Electrons from photosystem 2 go through ETC, producing ATP via chemiosmosis. Electrons from photosystem 1 reduce NADP+ to NADPH + H+. Electrons from photosystem 2 replace those lost in photosystem 1, electrons lost in photosystem 2 replaced by electrons generated from photolysis of water. Oxygen is by-product.
Cyclic (only photosystem 1 involved): High energy electrons released by photoactivation pass through ETC (producing ATP) and then return to photosystem 1. Electrons in cyclic photophosphorylation do not produce NADPH + H+, which is required for light independent reactions. Thus, though cyclic phosphorylation constantly creates ATP, the chemical energy cannot be used to create organic molecules.

Light independent reactions occur in the stroma and uses the ATP and NADPH + H+ from non-cyclic photophosphorylation. It is also known as the Calvin cycle and has three main steps:

Carbon Fixation:

Enzyme rubisco (RuBP carboxylase) catalyzes attachment of carbon dioxide to 5-carbon compound ribulose bisphosphate (RuBP). The new 6-carbon compound breaks down to two 3-carbon molecules: glycerinate-3-phosphate (G3P)
Reduction:

Each G3P molecule is phosphorylated by ATP and reduced by NADPH + H+, converting them to glyceraldehyde phosphate, a trios phosphate (TP)
Regeneration of RuBP:
For every 6 TP, only one can be used to create half a sugar molecule. The other 5 TP molecules are reorganized to regenerate RuBP (using ATP). The cycle repeats multiple times to construct chains of sugars (sucrose)

Draw the molecular structures of glucose, fructose and sucrose.

Distinguish between the structure and function of the three polysaccharides of glucose.

glycogen- multiple branches, animal storage of glucose
cellulose- branched and unbranched varieties, main structural component of plants
starch- branched, plant storage glucose

Compare the leaf and root structure and vascular tissue organization in stem and roots of monocotyledonous and dicotyledonous plants. [State which sugar cane is.]

Leaves: Monocots have parallel veins from stem to tip; dicots have branching veins
Root structure: Monocots do not have main roots but have smaller roots that branch out; dicots have main roots that grow downwards
Vascular tissue organization: Monocots have vascular bundles spread across the stem, with bundles closer to the sides of the stem rather than the center; dicots have vascular bundles circular

Task Two

State the meaning of 'allocation of photo-assimilates'.

The allocation/grouping of a certain amount of compounds formed through assimilation in light dependent reactions (monosaccharides) from a plants leaves to its sinks

Explain what the author means by the phrase 'heterotrophic sinks'.

The areas of a plant in which sucrose is stored: fruits, roots, etc.

Describe the structure and function of a plasmodesmata between the mesophyll and parenchyma cells.

Plasmodesmata are microscopic communication bridges between the cell walls of two plant cells; they facilitate the movement of sucrose from mesophyll cells to parenchyma cells

Hypothesize the effect of a pH or temperature change on plasmodesmata structure and function. [Hypotheses must include a why!]

A change in pH or temperature would denature proteins that make up the plasmodesmata, which would probably prevent it from functioning efficiently

Distinguish the structure and function of xylem and phloem.

Xylem vs. Phloem:

Structure: Non-living vs living; thin walled sieve tubes w/ pores at ends vs tubes with no cross walls
Function: Transports nutrients from leaves to sinks/growing parts vs water/minerals from roots to aerial parts (leaves)

Task Three

Distinguish between, in your own words, the apoplastic route and symplastic route of ion movement in plants.

Apoplastic route: through the cell wall
Symplastic route: through the cytoplasm (via plasmodesmata)

Draw the pathway of sucrose from mesophyll cell [site of production], to the SE/CC.

Draw and annotate a diagram to describe the pressure flow model of water and sugar transport in xylem and phloem. [direction and substance are a must!] [you may add onto the previous drawing if you like].

Annotate your drawing to distinguish between the different transport mechanisms [diffusion, osmosis, facilitative diffusion, primary active transport, secondary active transport ex. SUT1 ] that occurs in the xylem/phloem pressure flow model of sucrose and water transport.

Sugar, Insulin, Obesity and Cancer

noreply@blogger.com (Alex Chiang) — Sat, 03 May 2014 11:40:00 +0000

Task One [Monday Feb 17th]: Insulin Production:

Draw and annotate a pancreatic cell with all the structures and their functions for the production and export of proteins - specifically insulin.

Draw a cartoon image of insulin and.... Describe the different levels of protein structure and the forces at each level that maintain its share.

Task Two [Wednesday Feb 19th]: Insulin Physiology

Describe the normal cycle for insulin: stimulus, production cell, target cell, response, feedback.

Increase in level of glucose in blood leads to production of insulin in beta islets of pancreas
Insulin targets muscle (use glucose for ATP production), fat (energy storage), liver cells (glycogenesis)
This results in decrease of blood sugar levels in which less insulin is produced (negative feedback)

Distinguish between the storage of glucose in liver/skeletal muscle from storage in white adipose tissue [a connective tissue]

Liver/skeletal muscle: glucose to glycogen (glycogenesis)/glucose to ATP (metabolized)
Adipose tissue: glucose to fats

Distinguish between the three cases: normal insulin response, diabetes type 1 and diabetes type 2.

Normal insulin response: Insulin produced in beta islets, and binds to receptors of membranes on the target cells
Diabetes Type 1: cells body attack islet cells, inhibiting production of insulin
Diabetes Type 2: diet related, leading to receptors to be less receptive/sensitive to insulin

Task Three [Friday February 21st]: Response to Insulin - Cell Level

From the above quotes and reviewing the content of the core please answer the following questions:

Describe the target cell response mechanism in liver, skeletal and adipose tissues by:

Draw and describe the fluid mosaic model of the cell membrane

Predict the transport mechanism liver, skeletal and adipose tissues would use to absorb [uptake] glucose from the blood.

Active transport using a protein membrane due to large size of glucose

Compare and contrast the predicted mechanism to the actual process used by intestinal epithelium to absorb glucose from the intestinal lumen.

The mechanisms are similar in the intestinal epithelium (use GLUT2 as well)

Task Four [Tuesday, February 25th]: Response to Insulin - Gene Level

Explain the cancer cell dependency for glucose [energy and lipids]. There are two pieces of the mystery eluded to above:

One: Mitosis- cancer cells use glucose and glycogen for mitosis and growth

Two: Lipids- cancer cells use fat tissues to synthesize its own membrane

Cancer cells response to glucose requires more protein transporters in the cell membrane and more enzymes for glucose metabolism for mitosis, lipid synthesis and nucleotide synthesis.

Describe the entire process of gene transcription, mRNA production, [include splicing to remove introns], translation by ribosomes.

Gene transcription: RNA polymerase binds to promotor on DNA strand, unwinds strand, and uses one strand of the double helix as template for replication
mRNA production: RNA nucleotides brought by RNA polymerase, producing RNA strand; before leaving nucleus for ribosome, introns are cut out by spliceosomes in which a cap is added to one end and adenine tail is added to the other
Translation by ribosomes: Ribosomes bind to start codon of mRNA, which is followed by tRNA binding to the mRNA via the ribosome (tRNA is bound to amino acid); continuation of this leads to an amino acid chain. tRNA binds to mRNA in A site of ribosome, and is then moved to P site, which leads to another tRNA to bind to A site, forming peptide bond between amino acids of the two tRNAs, tRNA in P site is released in E site without amino acid.

Task Five [Thursday, February 27th]: Glucose metabolism

State and draw a basic cartoon of stages of glucose metabolism used to generate ATP in the non-proliferative tissues. [In class we broke the entire process into five steps]

Describe the input and output with regards to: NAD+, NADH+H, CO2, ATP, H2O

Distinguish between animal use of lactate fermentation and alcoholic fermentation by bacteria and yeast.

Yeast produces carbon dioxide and ethanol
Animals produce lactic acid

Explain the advantage of cells switching to anaerobic metabolism [fermentation] of sugars.

Anaerobic metabolism of sugars offers energy quicker because the process is faster

Distinguish between catabolic and anabolic reactions and the role of NAD+ and NADP+ in catabolism and anabolism.

Catabolic reaction: break down of molecules, giving out energy
Anabolic reaction: formation of molecules using energy; uses both NAD+ and NADP+

Review: Chemistry of Life/Recombination Genetics

noreply@blogger.com (Alex Chiang) — Mon, 28 Apr 2014 13:44:00 +0000

Task 1

State the essential amino acids for humans.

Tryptophan, Phenylalaline, Tyrosine

Explain the process of evolution that may have occurred to allow humans to survive without the de novo biochemical pathways to make these essential amino acids.

A mutation may have occurred early on before leading to genetic drift

Explain why ALL amino acids are non-essential for plants.

Plants cannot get adequate supplies of pre-formed amino acids so they have to make all of the amino acids needed for protein synthesis de novo.

State what makes the molecule chorismate important.

The three aromatic amino acids are all made from chorismate

Task 2

Explain how the types of inhibition that glyphosate COULD use to inhibit the EPSP enzyme.

Competitive inhibition: type of enzyme inhibition where binding of the inhibitor to the active site of the enzymes prevents binding of the substrate and vice versa
Noncompetitive inhibition: type of enzyme inhibition where inhibitor reduces activity of the enzyme and binds equally well to the enzyme whether or not it has already bound the substrate
Product inhibition: type of enzyme inhibition where the product of enzyme reaction binds to the enzyme and inhibits its activity

Task 3

State and describe the actual mechanism glyphosate uses to prevent the chromisate metabolism.

Competitive inhibition: type of enzyme inhibition where binding of the inhibitor to the active site of the enzymes prevents binding of the substrate and vice versa

Describe the model of enzyme activity discussed in this article.

Induced fit substrate binding: when enzyme binds to substrate and changes occur in the active site, enhancing catalysis, as enzyme converts substrate to product

Analyze the picture that shows the alanine --> glycine mutation. Explain the nucleotide changes that could have resulted in the alanine to glycine change. [there are many]

Deletion of nucleotide(s), frameshift after nucleotides were removed, or single base substitution

Explain the effect the G100-->A100 change has on the affinity of glyphosate for the PEP [substrate] active site.

The change may prevent PEP binding because the enzyme loses a molecule, changing the shape of the protein

Task 4:

Describe the IDEA of what Monsanto was attempting to do.

Monsanto was trying to genetically modify the plants via recombinant DNA to allow the plants to resist the Roundup herbicide

Explain, with proper sequence and language, the evolutionary process of this bacteria from this location.

"The C4 strain of Agrobacterium sp. proved to be just the thing. This is a species of bacteria that was found growing in the waste-fed column at a factory that made glyphosate. The EPSP synthase enzyme from this bacterium (C4 EPSP synthase) was almost completely insensitive to glyphosate."

The C4 strain of Agrobacterium sp. was originally susceptible to the glyphosate, but mutations occurred, which resulted in the production of a new kind of EPSP synthase enzyme that resisted the effects of glyphosate.

Explain the USE of the following components of the plasmid used by Monsanto:

ori-322 and ori-V - ori-322 is an origin of replication from plasmid pBr322 by E. coli; ori-v is an origin from plasmid RK2, a plasmid that reproduce in a wide variety of gram negative bateria

AAC(3)-III, along with P-35S and NOS 3' - AAC(3)-III gene allows the selection of gentamycin resistance in plants; P-35S is a promoter that indicates to ribosims where to begin replication; NOS 3’ is the stop codon

CP4-EPSPS along with P-e35S and NOS 3' and CTP2 - CP4-EPSPS gene was modified and a plant promoter (P-e35S) was added along with a polyadenylation site, NOS 3’. CTP2 is a N-terminal leader sequence that targets the protein to the chloroplast as EPSP synthase is in chloroplasts where synthesis of chorismate takes place.

Distinguish the following descriptions of transformation in plants to the process used to transform bacterial colonies. [pGlo experiment]

pGlo uses two main steps:

transformation solution containing CaCl2.

'heat shock' that induces gene expression that stimulate bacteria to 'take up' DNA from their environment.

Transformation of plant cells used:

"The interesting feature of this transformation is that it is mediated by the bacteria. All you need to do is expose the plant cells to the bacteria under the right conditions and your gene of interest will end up in a plant chromosome."

pGlo: plasmid heated to allow it to diffuse into bacteria
plant cells: no heat required, simple exposure will lead to bacteria to be in the plant chromosome

State the substance in the last image, that 'selects' the proper Agrobacteria with the genetically engineered plasmid to survive.

Glyphosate

State the plant hormones that are used to promote root and shoot development from the transformed leaf disk.

Auxin, Gibberellin, Abscisic acid

Observing Plant Roots and Fruits

noreply@blogger.com (Alex Chiang) — Fri, 14 Mar 2014 02:20:00 +0000

Apple

-Stem: Transport of fluids and nutrients to and from the fruit

-Exocarp: Tough outer skin to protect the fruit

-Mesocarp: The 'flesh' of the fruit, located between epicure and endocarp, is the edible part of the fruit that attracts animals

-Endocarp: Area that directly surrounds seeds, is thick and hard to protect the seed

-Seed: Carries, protects, and nourishes embryo until germination. It is necessary for the plants reproduction and survival (as a species)

-Hypanthium: Contains nectar

Bell Pepper

-Stem: Transport of fluids and nutrients to and from the fruit

-Exocarp: Tough outer skin to protect the fruit

-Mesocarp: The 'flesh' of the fruit, located between epicure and endocarp, is the edible part of the fruit that attracts animals

-Endocarp: Area that directly surrounds seeds, is thick and hard to protect the seed

-Seed: Carries, protects, and nourishes embryo until germination. It is necessary for the plants reproduction and survival (as a species)

-Placenta: Region on the inner fruit wall where seeds attach to

Eggplant

-Pedicel: Connects eggplant to rest of plant

-Calyx: Protect bud until it matures

-Exocarp: Tough outer skin to protect the fruit

-Mesocarp: The 'flesh' of the fruit, located between epicure and endocarp, is the edible part of the fruit that attracts animals

-Endocarp: Area that directly surrounds seeds, is thick and hard to protect the seed

-Seed: Carries, protects, and nourishes embryo until germination. It is necessary for the plants reproduction and survival (as a species)

-Placenta: Region on the inner fruit wall where seeds attach to

Carrot

-Storage root: Stores nutrients from soil

-Stem: Transport of fluids and nutrients to and from the fruit and connects carrot to rest of the plant

-Periderm: Protect interior tissues from weathering and destruction

-Xylem: Transports water and soluble minerals from roots to plant

-Phloem: Transports and distributes products of photosynthesis to plant

-Tap Root: Allows plants to reach further down the ground to maximize the access of water and nutrients from the soil

Cauliflower

-Cauliflowers are immature flower heads and are flower buds packed really tightly.

Strawberry

-Seeds on epidermis because they do not reproduce via seeds because they grow asexually through roots that grow and bear fruit.

Mark is in trouble

noreply@blogger.com (Alex Chiang) — Sat, 21 Sep 2013 10:09:00 +0000

Part I

1. What types of physiological problems do humans encounter at high altitudes?

Headaches, dizziness/lightheadedness, shortness of breath, fatigue

2. What symptoms did the climbers exhibit that might be related to altitude? Explain.

Headache, deep breathing, fatigue. Shortness of breath would be caused by the lack of pressure of oxygen in high altitudes which leads to the climbers to breath deeper. Fatigue is caused by the lack of oxygen as not enough oxygen is taken in (compared to sea level) to energize the cells. Headaches may also be caused by the lack of oxygen that is going to the brains.

3. Compare the air at 18,000 feet (atmospheric pressure 280 mm Hg) to the air at sea level (760 mm Hg). What specific changes in the primary atmospheric gases (nitrogen, oxygen, carbon dioxide) might occur? Are they significant?

The composition of earth's atmosphere is 78% nitrogen, 21% oxygen and 1% other gases (including carbon dioxide). There is less pressure at 18 thousand feet than there is at sea level, which means that the density of oxygen molecules is less and therefore would mean the amount of oxygen inhaled by a human being at 18 thousand feet would be lower than one at sea level. (Air pressure in atmosphere is supposed to be greater than the pressure inside the lungs so that the air would "go into" the lungs)

4. What is the specific pulmonary response to high altitude? [Assume you are considering a subject at rest.]

Lowered oxygen density in the atmosphere means less oxygen intake which means that the heart will pump harder to get more blood to pump throughout the body to try and catch up to the normal amount of oxygen that would be distributed at sea level.

5. How will this response affect overall blood gases? What about oxygen loading and unloading from hemoglobin? Explain how you arrived at your conclusions.

There will be less oxygen binding to the hemoglobin in the blood as there is less oxygen in the atmosphere. There will be a higher affinity of RBC for oxygen

6. After breathing at altitude for a few days, the body normally begins producing more 2,3-DPG. What is the significance of this change? How will it affect the pulmonary changes observed?

2,3 DPG helps decreases the affinity of RBC for oxygen which means that the body will be able to increase the amount of oxygen distributed to the tissues as less would stick to the RBC. The person will be less fatigued, have a lower heart rate and less headaches

Part II

1. What physiological changes is Emily referring to (above) that will occur when someone lives at altitude for an extended period?

The body learns to increase production of erythrocytes, which increases oxygen in blood as there are more RBCs.

2. How are these changes advantageous?

Cells in the body will receive more oxygen

3. What is the specific physiological pathway that results in the changes described?

Kidney detects low oxygen level and secrets erythropoietin, stimulates RBC production in bone marrow, more RBC = More hemoglobin = more oxygen in blood = more oxygen to tissues

Part III

1. How would the oxygen and Gammow bag help Mark?

Oxygen and Gammow bag will increase the amount of oxygen in Mark. The bag increases the atmospheric pressure, which leads to less strenuous breathing because more air is going into the lungs.

2. If you were a member of the medical team examining Mark, what types of tests would you run? Why? [Try to focus on what types of things you would like to measure, whether or not you know of a possible test for them.]

Blood sample, see concentration of oxygen/CO2 and pH levels

3. What types of results do you expect to find? Explain your reasoning.

Lower oxygen/CO2 levels in blood compared to sea level (lower pH) because there is less oxygen at 18 thousand feet than at sea level. So the body is trying to play "catch up" to maintain the distribution of oxygen but cannot.

Hominid Evolution and Carbon Dating

noreply@blogger.com (Alex Chiang) — Thu, 22 Aug 2013 10:11:00 +0000

Hominid Evolution

Ardipithecus ramidus (4.4 million years ago)

-divergent large toe with a rigid foot

-pelvis tree climbing and bipedal activity

-ape ancestor not chimpanzee like

-canine teeth = same size male and female

-wooded environment

-3 ft 11 in, 110 lbs

Australopithecus afarensis (2.95-3.85 million years ago)

-ape and human characteristics

-apelike features:

face proportions

braincase (small brain)

strong arms with curved fingers (climbing trees)

-human features:

small canine teeth

body stood on two legs

-could live on trees and ground

-males (4ft 11in, 92 lbs)

-females (3ft 5in, 64 lbs)

Australopithecus africanus (2.1-3.3 million years ago)

-rounder cranium w/ larger brain and smaller teeth

-apelike:

long arms

sloping face that juts out (pronounced jaw)

-pelvis, femur, footbones: walked bipedally

-shoulder and hand: climbing

-males (4ft 6in, 90 lbs)

-females (3ft 9in, 66 lbs)

Homo habilis (1.4-2.4 million years ago)

-larger braincase

-smaller face and teeth

-ape like features: long arms & prognathic face

-average (3ft 4in- 4ft 5in, 70 lbs)

Homo erectus (143k-1.89mil years ago)

-more adapted to the ground

shorter arms

long legs

=walk and run more

-average (4ft 9in - 6ft 1in, 88 - 150lbs)

Homo neanderthalensis (28-200 thousand years ago

-closest extinct human relative

-skull features

large middle part of face

angle cheek bones

huge nose for different air (humidifying and warming)

-shorter stockier bodies (cold environments)

-brain same size sometimes larger

males (5ft 5in, 143 lbs)

females (5ft 1in, 119lbs)

Homo sapiens (200,000 to present)

-large brains

-thin walled high vaulted skull

-flat almost vertical forehead

-smaller teeth

Summary

-The braincase gets larger incrementally

-Generally get taller and heavier

-Shorter arms and longer legs

-Nose smaller

-Smaller teeth

-Became only bipedal

-These traits are meant to suit us to live on the ground

http://humanorigins.si.edu/resources/intro-human-evolution

Carbon dating

Carbon dating is used to determine the age of fossils and requires the use of the radioactive isotope carbon-14. Carbon-14 is produced by cosmic ray protons which blast nuclei in the upper atmosphere, thus producing neutrons and bombard nitrogen creating carbon-14. This occurs at a rate which is constant, therefore we can use radioactive emissions of once-living matter and compare it to living ones. The results allows us to make a measurement of the time that passed.

Radioactive half-life is the time for half the radioactive nuclei to go through radioactive decay. Unstable radioisotopes decay into more stable forms (C-14 decays to C-12, which is stable.)

Living organisms constantly exchange carbon with the atmosphere in the form of CO2; this results with the organisms to have nearly the same ratio of C-14 to C-12 with the atmosphere. However when the organism dies, it beings to undergo radioactive decay in which C-14 decays into a more stable C-14. This can be used to measure the amount of time an organism had been dead for.

Carbon dating is used for younger fossils because of the difference in half-life.

C-14 has a half-life of 5700 years which is useful for fossils that are 1000 to 10000 years old, whereas potassium (K40) has a half life of around 1.3 billion years therefore allowing it to be used for much older fossils.

Abiogenesis and Panspermia

noreply@blogger.com (Alex Chiang) — Sun, 18 Aug 2013 12:18:00 +0000

The abiogenesis theory states that life came originated from simple organic compounds that were created on the earth from the pre-biotic soup that existed. However on the other hand, the panspermia theory states that life on Earth originated from comets and asteroids carrying bacterial life from the universe.

The Urey-Miller experiment provides support that the occurrence of abiogenesis is possible. The Urey-Miller experiment aimed to create organic compounds via a system that simulated similar conditions of pre-biotic Earth. The simulation consisted of a recirculating system with heated water (as the ocean), a chamber with electrical charges (as the atmosphere with lightning), and a condenser for gases in the 'atmosphere 'chamber to dissolve in the water. These chambers consisted of ammonia, nitrogen, methane, hydrogen gas, and carbon dioxide. After a while, the experiment's result showed simple organic molecules in the 'atmosphere' chamber, whereas amino acids and other organic acids were found in the 'ocean' chamber.

However, it is still possible for the panspermia to be valid because abiogenesis may have occurred on a different planet and comets may have transported the products to Earth while abiogenesis was occurring on Earth.

Scientists think the RNA replication or metabolism occurred first before protobiont cells began to form in micro environments for many reasons.

Scientists think RNA replication occurred first because RNA have the ability to replicate themselves as well as catalyzing reactions. RNA also have the ability to attract and link amino acids into proteins, which are the building blocks of life. Furthermore, scientists believe that RNA were able to persist because of the clay from volcanic ashes would cover it from harmful UV rays.

Scientists also argue that metabolism occurred before the protobiont cells began to form because they believe that the Earth may have started out with the first living thing on Earth as a primitive metabolic life sustaining itself with a series of reactions with CO2 or CO catalyzed by metal sulfide.

Protobionts are thought to be precursors to prokaryotic cells. They consist of abiotically produced organic molecules surrounded by a membrane like structure. Protobionts have characteristics that are related with life such as simple reproduction, metabolism, excitability, and the maintenance of itself to survive in its environment (with the help of its membrane). Due to this, many scientists believe RNA replication and metabolism occurred before protobiont cells began to form. RNA is needed for the protobiont cells to execute simple reproduction and metabolism is needed so it can energize itself to survive.

However, whether RNA replication or metabolism may have been the very first occurrences of life, both the abiogenesis and panspermia theories still have validity because organic compounds could have still been created on Earth or survived on a comet traveling to Earth to create protobiont cells.

Pre-lab

noreply@blogger.com (Alex Chiang) — Tue, 21 May 2013 12:14:00 +0000

Outline a basic technique used for gene transfer involving plasmids, a host cell (bacterium, yeast or other cell), restriction enzymes (endonucleases) and DNA ligase.

Gene transfer involving plasmids are usually used to alter the circular DNA molecules called plasmids that are usually found in plasmids.

Stage 1: Obtaining a gene to transfer

-Restriction enzymes are put in the same container as the DNA strands

-During this stage, restriction enzymes are used to cut the targeted gene that is to be transferred from the DNA of an organism to the plasmid and create 'sticky ends'

Stage 2: Preparing a vector for the transferred gene

-Restriction enzymes are also put in the same container as the plasmids

-The same restriction enzymes that had cut the targeted gene from the DNA strand is used to create a cut in the plasmid, creating 'sticky ends' complementary to the DNA fragments from an organism

Stage 3: Recombinant DNA

-DNA ligase catalyzes the reaction and reconnects the cleaved plasmids with the DNA strands by creating covalent bonds

Stage 4: Isolation of transformed cells

-The recombinant DNA is then inserted into the host cell

-The cells that are transformed to contain the recombinant DNA are separated from untransformed cells

Crime Scene Investigation

noreply@blogger.com (Alex Chiang) — Tue, 21 May 2013 12:13:00 +0000

1) Explain the content of the human genome: % protein encoding, % regulatory, % STRs AND distinguish locus from gene from alleles

-There is less than 2% protein encoding (code for proteins) in the human genome, around 3% STRs (short sequences of non-coding DNA), and around 8% regulatory genes that control gene expression

http://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Components_of_the_Human_Genome.jpg/370px-Components_of_the_Human_Genome.jpg

-A locus is the position of a gene on a chromosome

-A gene is the specific part of DNA base pairs on a chromosome that code for a phenotype

-An allele is a different type of gene in the same locus

http://www.csulb.edu/~kmacd/361-6-Ch1_files/allele.jpg

2) Explain the use of STRs in profiling to demonstrate the rarity of sharing all/most/many STRs

STRs (short tandem repeats) are used in DNA profiling because they can be highly polymorphic, which means there are many varieties of it. Therefore, STRs are very unique to each individual and therefore it can be used to discriminate between individuals esp. at a crime scene.

3) State and describe the evidence of the gel to identify the suspect at the crime scene by referring the alleles found at the crime scene and the allele of the suspect(s)

-On the gel, the second column is the DNA found in the crime scene, which is subsequently followed by Suspect A in column three, Suspect B, Suspect C and Suspect D in the next three columns.

-As you can see below, the DNA found at the crime scene is exactly in line with Suspect C of the fourth column. This is because the alleles (one blue dash is an allele) exactly correspond

DNA Profiling

noreply@blogger.com (Alex Chiang) — Sat, 11 May 2013 07:25:00 +0000

Write a brief description of the difference between RFLP and STRs

-RFLP (restriction fragment length polymorphism)- it indicates a difference between samples of homologous DNA molecules via restriction enzymes

-STRs (short tandem repeats)- they consist of short repeated sequences of DNA (2-8 nucleotides) and are located on non-coding regions of chromosomes. STRs are also polymorphic, which means that there are nine or more different types of alleles that can be detected.

-The STR and RFLP methods differ in DNA profiling and the STR has more advantages, which is why it is preferred today. While the STR only needs a very small amount of DNA to run, the RFLP does not as it requires a very large sample of DNA to work. Therefore, with a small amount of DNA, using the STR method reduces the likelihood of destroying sample due to the utilization of PCR. On the other hand, the RFLP method destroys the cell after it is completed. Furthermore, the STR method takes only around a couple hours to complete whereas the RFLP method could take weeks. STR is also more accurate than RFLP because STR targets specific sites unlike RFLP.

How many STRs does the FBI use to analyze?

-They use 13 different STRs, with repeats of four or five nucleotides, plus sex markers

Describe how PCR is used in DNA profiling

-PCR is used to amplify the specific regions of the DNA (STRs) and is therefore used to ultimately detect differences between STRs. During PCR, DNA fragments, with primers, are heated and cooled to activate primers and DNA polymerase in copying the fragments over and over again until the amount of copied fragments are "amplified"

Describe how electrophoresis is used in DNA profiling

-Electrophoresis is used to separate different DNA fragments (which were amplified with PCR) by putting the fragments onto polyacrylamide gel. The polyacrylamide gel in PCR separates different DNA fragments by size and the smaller the fragment, the further it travels through the gel.

Explain how the probabilities of having a specific set of STRs is extremely unlikely. Therefore if yours is present at a crime scene, what does that mean?

-The probability of having a specific set of STRs is extremely unlikely because one in every 10^-100 quadrillion share the same STR profile. The number is greater than the number of humans that ever lived, therefore each STR profile is specific to one person. If my STR is present at a crime scene, this means that you were definitely at the crime scene but does not guarantee any more than that.

Recombinant DNA Techniques

noreply@blogger.com (Alex Chiang) — Tue, 07 May 2013 13:33:00 +0000

Extracting DNA

Reasons to Collect DNA

-Genetic testing

-Body identification

-Analysis of forensic information

Outline and explain the purpose of: swabbing, lysis buffer, centrifugation, and isopropyl alcohol

-Swabbing is the process of collecting a sample of cells. An example of this is swabbing the inside of somebody’s mouth to collect cells for DNA extraction.

-The lysis buffer contains detergent and enzyme proteinase K and separates the DNA from the cell. The detergent breaks down the cell membrane and nuclear membrane so that the cells would burst open and release the DNA. The enzyme cuts down the histone proteins, which the DNA is wrapped around, and therefore frees the DNA.

-Centrifugation causes the remaining debris and protein to sink to the bottom of the test tube while the DNA remains distributed in the liquid. This thus separates the DNA from any debris and protein, which may interfere with analysis.

-Centrifugation can also be used after isopropyl alcohol is mixed with the DNA so that the DNA would sink to the bottom of the tube, allowing the DNA to be isolated.

-Isopropyl alcohol concentrates the DNA together so that it is visible to the naked eye. This is because DNA is not soluble in isopropyl alcohol.

Cutting DNA

Explain how enzymes cut DNA

-Enzymes that cut DNA are called restriction enzymes and they are mostly found in bacteria. These enzymes are sequence specific, which means that they only bind and cut specific DNA sequences. After the restriction enzymes bind to a specifically recognized sequence, they cut down the sugar phosphate backbones of the DNA strands.

Find and compare 3 different enzymes (one which must cut with ‘sticky ends’ and another with ‘blunt ends’)

-AluI (source: Arthrobacter luteus) produces blunt ends

-BamHI (source: Bacillus amyloliquefaciens) produces sticky ends

-EcoRI (source: Escherichia coli) produces sticky ends

-AluI, which produces blunt ends, cuts the strand symmetrically whereas BamHI and EcoRI, producing sticky ends, cut the strand asymmetrically. This means that BamHI and EcoRI create 5’ or 3’ overhangs on the DNA strand, which makes it easy to ligate. On the other hand, enzymes such as AluI leave no overhang because it cuts the strand symmetrically.

Separating nucleic acids/protein

Outline and explain steps: purpose of gel, electricity, ‘running buffer’, DNA standard, ‘loading buffer’, ethydium bromide

-The purpose of gel is to sort out DNA strand by filtering them with its sponge-like consistency.

-The electricity during electrophoresis makes the DNA strands in the wells of the gel move through the gel via a current. The smaller the strand, the further it will go in the gel.

-Running buffer is mixed with agarose when making the gel so that electrical charges can flow through the gel

-DNA standard contains DNA strands with known lengths; therefore it is used as a reference to compare the strands from the DNA sample with those from the standard.

-A loading buffer contains a dye that makes the DNA sample visible as it goes through the gel. Furthermore it makes the DNA sample thicker so that it will stay in the gel well instead of floating away.

-Ethydium bromide is used to stain the DNA sample because it binds to the DNA and is visible under fluorescent light.

Amplifying DNA

Outline and explain steps: role of primers, heating, cooling, reheating, over and over, compare amount of DNA at start to amount of DNA at finish.

-The primers in PCR attach to sites on the DNA strand that are at either end of the segment that is targeted to be copied.

-Heating the DNA causes it to unwind into two separate strands. Cooling it then leads to the primers in the solution to attach to the targets before the two DNA strands can rejoin. The solution gets heated again, causing the DNA polymerase to attach on a primer and add complementary nucleotides onto the strand. The temperature is then heated, cooled, and heated again to repeat this cycle, creating more complementary strands of the target sequence. After 4 cycles, there are 8 fragments that contain only the target sequence. After 30 cycles, there are over a billion fragments that contain only the target sequence and 60 copies of the longer length molecules.

April 25 Reflection

noreply@blogger.com (Alex Chiang) — Thu, 25 Apr 2013 05:39:00 +0000

Question

Do the differences in the amino acid sequences of the cytochrome c protein affect the morphological and genetic difference between organisms

Claim

The differences in amino acid sequences of conservative proteins such as cytochrome c affects morphological and genetic differences between organisms.

Evidence

-As you go down the cladogram from one organisms, the other organisms have increasingly more differences in the amino acid sequence.

-The difference in the amino acid sequence between a rhesus monkey and human is 1

-The difference in the amino acid sequence between a tuna and human is 20

Justification

The rhesus monkey and humans share more morphological and genetic similarities than the tuna and human. This correlates with the difference in the amino acid sequence for cytochrome c.

Being a presenter was not a very special thing because it was simply showing ideas.

Being an interviewer is more interesting because you get to act stupid

I think I could present the information somewhat smoothly but not perfectly

I feel I could justify the claim using the evidence

I stumbled occasionally in the explanation

Cambrian Shutter of Doom Becomes Sucker of Worms

noreply@blogger.com (Alex Chiang) — Thu, 21 Mar 2013 13:45:00 +0000

Summary

This article talks about the discovery of fossils of Anomalocaris, a Cambrian segmented invertebrate. It primarily concentrates on the shape of their bodies based on fossil evidence, which would impact the assumptions on the creature. The article presents the different hypotheses that paleontologists had on the structure of Anomalocaris. It begins by introducing paleontologist Charles Walcott’s idea that it was an archaic form of a jellyfish. However, the discovery of their mouth structures gave evidence that Anomalocaris was a predator that fed on hard-shelled trilobite. This was, again, contradicted by evidence that claimed Anomalocaris didn’t feed on hard-shelled trilobites due to their mouth shapes. It is then presented that the Anomalocaris probably fed on small prey in the mud by sucking them up instead of chomping on them.

Fossils used to determine age of organism

The age of a fossil can determine the age of the organism and seeing which layer of rock it is found in can identify the fossil’s age. The older the fossil, the deeper it would be located in the layers. According to the article, scientists found out the age of the Anomalocaris fossil by identifying other fossils found nearby in the same site (Burgess Shale Site.) The article reveals that the fossils were 505 million years old.

Morphology providing evidence for common ancestry

Anomalocaris was such a strange creature to the point where scientists assumed the parts that made up the creature were from different creatures and it was only recently that scientists found out what it looked like. The creature is described to have stalked eyes, swim flaps, spiked grasping appendages, and plated mouths making it look alien-like. When scientists assumed it was an archaic jellyfish, this was primarily due to a flattened ring that was identified on the fossil, which were the squishy discs on the mouth of the Anomalocaris. This misconception, however, was replaced with the discovery of its different body parts that made it become categorized with arthropods in the fossils. This is due to Anomalocaris’s segmented body that its swim flaps extend out from. This means that current arthropods such as shrimps have a high chance that they evolved from Anomalocaris.

http://www.wired.com/images_blogs/wiredscience/2011/05/Image-1-Laggania-660x440.jpg

What it is

The four major evolutionary developments start with the development of three tissue layers, continuing with the development of, in order, bilateral symmetry, internal cavity, and type of zygote cleavage. The three tissue layers that are developed in triploblastic creatures such as arthropods are the ectoderm, endoderm, and mesoderm. However, tissues were never brought up in the article so there is no way to determine if Anomalocaris are animals through this development. Despite this, we can see from the picture above that Anomalocaris is an animal because it has a bilateral symmetry. Each half of Anomalocaris has an eye, a spikey appendage, and swim flaps. Thus, this gives support to the argument that Anomalocaris are animals. It is implied in the article that Anomalocaris have internal cavities simply because it discusses about the mouth of the creature and what it probably fed on (hard-shelled trilobites or scraps on the ocean floor such as worms.) This means that Anomalocaris is either a pseudocoelomate or a coelomate. Psuedocoelomates and coelomates have body cavities unlike acoelomates but psuedocoelomates develop their body cavities between the endoderm and mesoderm whereas coelomates develop their coeloms in the mesoderm. According to the article, it is not mentioned whether the Anomalocaris is a psuedocoelomate or a coelomate. It is known that the type of zygote cleavage developed primarily applies to coelomates in which they can be divided into two groups of protostomes and deuterostomes. In protostomes, the zygote divides into two cells, which keep dividing and end up forming the blastula. The blastopore, which is the region where initial opening occurs during the early stages of embryonic development, forms to become the mouth in protostomes. While in deuterostomes, the blastopore becomes the anus first. The article, however, does not discuss anything related to the type of zygote cleavage that Anomalocaris could have possibly developed. Based on this evidence alone, we can identify that the Anomalocaris was an animal that had bi-laterally symmetrical bodies and, in addition, was either a pseudocoelomate or coelomate.

What it isn’t

It is not radially symmetrical because if it was sliced in several different ways, there wouldn’t be any matching sections. It is also not an acoelomate because they although they are bilaterally symmetrical like acoelomates, they have body cavities. Therefore we can rule out all claims that it was either Ctenophoras, Cnidarias, Poriferas, or Choanoflagellates as these are all organisms that do not have bilateral symmetrical bodies. Anomalocaris also are not Platyhelminthes or Acoelomorphas because these are organisms that do not have body cavities.

What is not preserved in a fossil sample

Tissue is not preserved in fossil samples because it does not preserve well, especially compared to bones. Therefore, it is hard for scientists to determine a creature’s tissue sample based on fossil evidence. This is probably why the article did not discuss issues such as whether Anomalocaris were pseudocoelomate or coelomates, which in turn would make it hard to determine whether it was a protostome or deuterostome. These facts can only be hypothesized based on evidence from modern world descendants of the Anomalocaris such as shrimps.

Source: http://www.wired.com/wiredscience/2012/05/cambrian-shutter-of-doom-becomes-sucker-of-worms/

Cladogram

noreply@blogger.com (Alex Chiang) — Tue, 12 Mar 2013 15:16:00 +0000

My Cladogram

Reasoning

Frog:

Frog Characteristics according to dissection:

-It has a skeleton with a spinal cord as provided in the second picture

-It has paired appendages (paired eyes, legs, organs)

-It has a visible digestive tract

-The reason why I think it is the most complex is because it has legs in addition to it's existing traits

Fish:

Fish Characteristics according to dissection:

-It has a skeleton with a spinal cord

-It has paired appendages (fins, eyes)

-It has a visible digestive tract

-However I do not believe it is not more complex than a frog because it does not have legs

-Despite this, I believe the fish is more complex than the eel because it has more fins

Eel:

digestive tract

Eel Characteristics according to dissection:

-It has a skeleton with a spinal cord

-It has paired appendages (fins, eyes)

-It has a visible digestive tract

-However I feel it is less complex because it has less fins than a fish (it only has dorsal, anal, and caudal fins)

-It is more complex than a starfish because it has a spinal cord

Starfish:

Starfish Characteristics according to dissection:

-It has a skeleton, but with no spinal cord

-This is due to the fact that is has no paired appendages

-The reason the starfish is more complex is because it has a visible digestive tract (seen in the centre which is it's stomach)

Crab:

Crab Characteristics according to dissection:

-It has an exoskeleton

-Crabs have paired appendages (legs, eyes, claws)

-A digestive tract is also visible after cracking it open

-I believe it is more complex than a grasshopper because it has more internal organs (such as gills, visible heart, more muscles and tissues)

Grasshopper:

Grasshopper Characteristics according to dissection:

-The grasshopper has an exoskeleton

-It also has paired appendages (legs, eyes, wings)

-I am assuming that the organ sticking out (near the bottom-left of the picture) is part of the grasshopper's digestive tract (the stomach)

Shrimp:

Shrimp Characteristics according to dissection:

-It has an exoskeleton

-It has paired appendages (eyes, legs, swimmerets, antennae)

-However according to the pictures, there is no visible digestive tract (even though my experiences of eating shrimps has led me to discover a digestive tract)

-If there was a visible digestive tract, which there should be, I would place the shrimp right before the crab on the cladogram because crabs have specialized appendages (for swimming) and because the grasshopper has less internal organs. Crabs and Shrimps are also both aquatic organisms, which is also why I think they are more closely related

Clam:

Clam Characteristics according to dissection:

-It has no skeleton but is protected by a shell

-It also has no appendages

-There is, however, a visible digestive tract

-The reason why I believe a clam is more complex than a worm is because of the shell to protect it

Worm:

Worm Characteristics according to dissection:

-It has no skeleton

-Nor does it have any visible appendages

-It has a digestive tract

-However, I believe it is the least complex because it seems to have the least amount of protection in the case that is has no skeleton or shell to protect itself

October 15 Homework

noreply@blogger.com (Alex Chiang) — Sat, 13 Oct 2012 16:19:00 +0000

2.2.1 - 2.2.2 (Label and annotate a prokaryotic cell)

http://biology.about.com/od/cellanatomy/ss/prokaryotes.htm

2.2.3 (Identify prokaryotic structures in electron micrographs)

http://sociallyconsciousbird.com/storage/hlbio/cellbionotes_files/image002.jpg

2.3.1 - 2.3.2 (Label and annotate an animal cell)

http://biology.about.com/od/cellbiology/ss/animal_cells.htm

2.3.3 (Identify animal cell structures in electron micrographs)

http://www.ibguides.com/images/biology/2.3.2.png

2.3.5 (State three differences between plant and animal cells)

-Plant cells have both vacuoles and cytoplasm while animal cells only have cytoplasm

-Plant cells have both cell walls and membranes while animal cells only have cell membranes

-Plant cells have chloroplast while animal cells do not

http://library.thinkquest.org/C004535/media/eukaryote_plant.gif

Osmosis Lab (IA #2)

noreply@blogger.com (Alex Chiang) — Sat, 13 Oct 2012 13:37:00 +0000

Graph did not show up above due to technical issues
(Blue vertical lines - Error bars for 4% sucrose solution)
(Red/Pink vertical lines - Error bars for 2% sucrose solution)