BIOL107 Lecture Notes - Lecture 3: Lipid Bilayer, Unsaturated Hydrocarbon, Alkane

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pinkworm218

17 Oct 2017

School

University of Alberta

Department

Biology (Biological Sciences)

Course

BIOL107

Professor

michaelharrington

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BIOL107 Full Course Notes

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Diplococcus bacteria (2 cells), they reproduce to make 2 paired cells. Small cell vs large cells: glucose get transferred into the cell through the membrane. Large cells cannot produce enough nutrients from the glucose but small cells are able to produce enough nutrients. Large organisms are made up of many small cells because it will increase the surface-to-volume ratio and also is beneficial for cell specialization. Large cells produce nitrogen (n2) which can be used to make amino acids and are linked to smaller paired cells, which they can help produce enough nutrients for the large cells through photosynthesis (co2 sugars) Separate the interior of the cell from the outside environment and contain contents of the cell. There are no covalent bonds within the phospholipid bilayer, they are bonded together from their (cid:862)dislike of (cid:449)ater(cid:863), so it is a flexi(cid:271)le material.

Related Questions

20. All living things must maintain a constantinternal environment to function properly through the process of__. (Points : 2)

metamorphosis
alkalosis
homeostasis
cryogenesis

Question 21.21. Glucose, fructose, sucrose,starch and glycogen are all examples of ___. (Points : 2)

proteins
carbohydrates
lipids
nucleic acids

Question 22.22. Some insects can "walk onwater" because __. (Points : 2)

water has a high surfacetension
water is adhesive
water istransparent
water expands when itfreezes

Question 23.23. To overcome an energy barrierbetween reactants and products in a chemical reaction, energy mustbe provided to get the reaction started. This energy is called ___energy. (Points : 2)

potential
reaction
initiation
activation

Question 24.24. Where in the cell doesglycolysis takes place? (Points : 2)

lysosomes
cytoplasm
mitochondria
nucleus

Question 25.25. Photosynthesis uses sunlightto convert water and carbon dioxide into __. (Points : 2)

carbon dioxide andwater
oxygen and sugars
proteins and sugars
carbon dioxide andoxygen

Question 26.26. The _____ of photosynthesisare the _____ of cellular respiration. (Points : 2)

electrons, protons
pigments, products
products, reactants
elements, atoms

Question 27.27. The energy currency of thecell is __. (Points : 2)

glucose
ATP
starch
glycogen

Question 28.28. When blood glucose levelsdecrease (as between meals), what reserves are tapped first?(Points : 2)

glycogen
fats
steroids
amino acids

Question 29.29. When glucose is used as anenergy source, the largest amount of ATP is produced by the ____portion of the entire respiratory process. (Points : 2)

glycolytic
Kreb's cycle
acetyl-CoAformation
electron transportchain

Question 30.30. Pyruvic acid can be regardedas the end product of ____. (Points : 2)

glycolysis
fermentation
the Kreb cycle
the electron transportchain

Question 31.31. How can one increase the rateof a chemical reaction? (Points : 2)

Increase the activationenergy.
Cool the reactants.
Decrease the pH.
Add an enzyme(catalyst).

Question 32.32. A prokaryote has ___. (Points: 2)

membrane-boundorganelles
nucleus
mitochondria
a cell membrane

Question 33.33. Unlike animal cells, plantcells have __. (Points : 2)

cell membranes
cell walls
chromosomes
mitochondria

Question 34.34. Which of the followingrequires energy to move molecules across the cell membrane? (Points: 2)

diffusion
osmosis
active transport
passive transport

Question 35.35. The ___ is the site of proteinsynthesis. (Points : 2)

nucleus
ribosome
mitochondrion
cytoskeleton

Question 36.36. Diffusion of water through asemipermeable membrane is called __. (Points : 2)

osmosis
diffusion
active transport
passive transport

Question 37.37. Which component of a cellmembrane is likely to function as a transporter? (Points : 2)

cholesterol
lipids
proteins
carbohydrates

Question 38.38. If potato slice is placed indistilled water, the cells would most likely __. (Points : 2)

lose water and becomelimp
neither take up water orlose water
take up water andswell
transport water into thecell by active transport.

Question 39.39. The chloroplast functions in___. (Points : 2)

protein synthesis
lipid storage
DNA replication
photosynthesis

Question 40.40. The fluid mosaic modelpresents the modern view of ___. (Points : 2)

the cell membranestructure
chromosome movementduring mitosis
protein synthesis
ATP production

Question 41.41. During which stage of mitosisdo nucleoli reappear and the nuclear membrane reforms? (Points :2)

prophase
metaphase
anaphase
telophase

Question 42.42. During which stage of mitosisdo chromatids separate to form two sets of daughter chromosomes?(Points : 2)

prophase
metaphase
anaphase
telophase

Question 43.43. Which of the following is thelongest stage of mitosis? (Points : 2)

prophase
metaphase
anaphase
telophase

Question 44.44. When during the cell cycle isDNA replicated? (Points : 2)

metaphase
Synthesis phase (S)
G1
cytokinesis

Question 45.45. A cell with 10 chromosomesundergoes mitosis and cell division. How many daughter cells areproduced and what number of chromosomes do they have? (Points :2)

4 daughter cells, 20chromosomes each
4 daughter cells, 5chromosomes each
2 daughter cells, 10chromosomes each
2 daughter cells, 5chromosomes each

Question 46.46. Maize (corn) has a haploidnumber of 10. How many chromosomes are in a zygote of maize?(Points : 2)

5
10
20
Not enough informationprovided

Question 47.47. Crossing over is one of themost important events in meiosis because __. (Points : 2)

homologous chromosomesare separated into different daughter cells
the number of chromosomesallotted to each daughter cell must be halved
sister chromatids must beseparated into different daughter cells
it produces new arrays ofalleles on chromosomes and contributes to genetic diversity

Question 48.48. While viewing stained onionroot tip cells you notice that distinct chromosomes cannot beobserved in most of the cells. This is because most of the cells___. (Points : 2)

were killed whilepreparing the slide
are in Interphase
are in metaphase
are undergoingMeiosis

Question 49.49. Each offspring cell producedby binary fission contains __. (Points : 2)

half the chromosomes ofthe original cell
twice as many chromosomesas the original cell
an independent assortmentof the original cell's chromosomes
an identical copy of theoriginal cell's chromosome.

Question 50.50. A DNA strand has the sequenceA-C-A-G-C-C-G-T-A. What would be its complementary strand? (Points: 2)

G-T-G-A-T-T-A-C-G
A-C-A-G-C-C-G-T-A
T-G-T-C-G-G-C-A-T
U-G-U-C-G-G-C-A-U

erinmoose680

Question 1

Which of the following provides the best description of adifference between prokaryotic and eukaryotic cells?.l.14.3>

The cell membrane of prokaryotic cells contains proteins, whilethe cell membrane of eukaryotic cells contains only lipids andcarbohydrates.

The DNA in eukaryotic cells is enclosed in the nucleus, whilethe DNA in prokaryotic cells is not separated from the rest of thecellÃ¢â¬â¢s contents.

Eukaryotic cells have cell walls that consist mainly of sugars,while prokaryotic cells do not.

Prokaryotic cells contain chloroplasts and mitochondria, whileeukaryotic cells do not contain specialized organelles.

Question 2

Cellular respiration is a process involving the breakdown ofglucose molecules in order to generate energy for cellularactivities. Given the general reaction below, which of thefollowing best describes the products labeled x andy?

C₆H₁₂O₆ + O₂ Ã¢â â _x___+ __y___ + energy.l.18.8>

The products are CO (carbon monoxide) andH₂O₂ (hydrogen peroxide), which thendecompose into carbon dioxide (CO₂) and water(H₂O).

The products are H2O (water) and CO₂ (carbondioxide), which then undergo anaerobic respiration.

The products are CO₂ (carbon dioxide) andH₂O (water), which then become the reactants forphotosynthesis.

The products are O₂ (oxygen) and H₂O(water), which are then used for growth and to keep the organismhydrated.

Question 3

If a researcher added a quantity of phospholipid molecules to abeaker of water, what would be most likely to happen to theorientation of the phospholipid molecules?.l.14.2>

They would break apart, and the remaining lipid tails wouldfloat up and form a layer on the surface of the water.

Their lipid tails would align within a spherical double layerwith their phosphate heads in the water inside and outside thesphere.

They would form a single-layered sphere with the phosphate headspointing outward and the lipid tails pointing inward.

Their lipid tails would adhere to the glass of the beaker andtheir phosphate heads would float in the water.

Question 4

While looking at a cell under a microscope, a scientist is ableto see a biological molecule. This molecule is a nucleic acid witha ladder structure and contains the nitrogenous bases adenine,thymine, cytosine, and guanine. Based on this information, whatconclusion can the scientist draw regarding this nucleicacid?.l.18.1>

The nucleic acid is RNA and would be found in the cellÃ¢â¬â¢snucleus.

The nucleic acid is RNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cellÃ¢â¬â¢snucleus.

Question 5

Which of the following is needed for cellular respiration tooccur?.l.18.8>

Sodium

Oxygen

Water

Nitrogen

Question 6

What would be the most likely effect if the side chain of an aminoacid group was unable to vary its structure?.l.18.4>

The organism would not be able to make and store starches.

The organism would not be able to process glucose moleculeseffectively.

The organism would not be able to regulate the creation ofspecialized cells.

The organism would not be able to make more than one type ofprotein.

Question 7

During photosynthesis, unused oxygen atoms form oxygen gas(O₂) which is released as a waste product. How did theseoxygen atoms originally enter the plant?.l.18.7>

They were part of the ATP molecules gathered by thechloroplasts.

They were part of the sugar molecules(C₆H₁₂O₆) the plant brought in forfood.

They were part of the carbon dioxide (CO₂) moleculesthat entered through the leaves.

They were part of the water molecules (H₂O) pulled inthrough the roots.

Question 8

Which of the following organelles houses DNA in eukaryoticcells?.l.14.3>

Endoplasmic reticulum

Golgi apparatus

Nucleus

Mitochondrium

Question 9

M. J. Schleiden shifted the scientific community's attention tocellular processes with his.l.14.1>

attention to disproving spontaneous generation

publication of ideas on cell generation

theories on the division of cells

claim that cells were the basic unit of all plant matter

Question 10

While the average human is able to hold his or her breath forapproximately one minute, a whale can dive for over 30 minuteswithout returning to the surface. Which of the following correctlydescribes this difference?.l.18.10>

Whales need less energy than humans.

Whales gather energy from their environment better thanhumans.

Whales are able to tolerate the products of cellular respirationbetter than humans.

Whales have a slower cellular respiration metabolism rate thanhumans.

Question 11

What can you conclude about the advantage of the cell membranehaving hydrophobic ends in the phospholipid bilayer?.l.18.3>

The cell requires the ability to digest and eliminateenzymes.

The cell requires a barrier between itself and itsenvironment.

The cell requires a means to store starches and glucose.

The cell requires a structure for eliminating waste.

Question 12

What would be the most likely outcome if the ribosomes of a cellwere damaged and stopped functioning?.l.14.3>

The cell would lose structure and shape.

Cell growth and repair would slow or stop.

Cell waste would no longer be processed.

The cell would produce more mitochondria.

ivoryjackal958

HELP ASAP; Please help me understand my lab results, so be detailed to clearly understand

Then , Please include answers to the question below to include in your response

1. what role does ampicilin (amp) and arabinose (ara) play in the experiment.

2) Because GFP was isolated from jelly fisht, what modification to the GFP DNA had to be made to result in expression in E. coli?

BACKGROUND INFORMATION

Transformation of GFP and the pGLO Plasmid GFP is a commonly used reporter protein in research labs, as the fluorescence creates a marker protein that can be used in many types of cell biology and biochemical studies. In basic research, GFP is often fused to a specific target protein of interest, creating a chimeric reporter protein. GFP has been used as a reporter protein to study blood vessel and tumor progression in mice, brain activity in mice, and malaria eradication in mosquitoes for instance (see website mentioned in notes). In the first week you will use a procedure to genetically transform bacteria with the gene that codes for Green Fluorescent Protein. Following the transformation procedure, the bacteria express their newly acquired jellyfish gene and produce the fluorescent protein, which causes them to glow a brilliant green color under ultraviolet light. During the second week you will purify the protein away from all of the other proteins produced by the bacteria using chromatography, and during the third week you will evaluate the success of the purification procedure and estimate the size (molecular weight) of GFP by using sodium dodecylsulfate polyacrylamide gel electrophoresis (SDS-PAGE). Fundamental to the process of genetic transformation of bacteria are plasmids. In addition to one large chromosome, bacteria naturally contain one or more small circular pieces of DNA called plasmids. Plasmid DNA usually contains genes for one or more traits that may be beneficial to bacterial survival. In nature, bacteria can transfer plasmids back and forth allowing them to share these beneficial genes. This natural mechanism allows bacteria to adapt to new environments. The occurrence of bacterial resistance to antibiotics is due in large part to the transmission of plasmids which contain antibiotic resistance genes. The pGLO plasmid contains the gene for GFP, and also contains the gene for the enzyme Î²-lactamase (bla), which provides resistance to the antibiotic ampicillin. The Î²-lactamase protein is produced and secreted by bacteria that contain the plasmid. Î²-Lactamase inactivates the ampicillin present in the LB nutrient agar, allowing bacterial growth. Only transformed bacteria that contain the plasmid and express Î²-lactamase can survive on plates that contain ampicillin. Only a very small percentage of the cells take up the plasmid DNA and are transformed. Untransformed cells cannot grow on the ampicillin selection plates. pGLO also incorporates a special gene regulation system, which can be used to control expression of the fluorescent protein in transformed cells. The gene for GFP can be switched on in transformed cells by adding the sugar arabinose to the cellsâ nutrient medium. Selection for cells that have been transformed with pGLO DNA is accomplished by growth on antibiotic plates. Transformed cells will appear white (wild-type phenotype) on plates not containing arabinose, and fluorescent green when arabinose is included in the nutrient agar medium.

The transformation procedure involves three main steps. These steps are intended to introduce the plasmid DNA into the E. coli cells and provide an environment for the cells to express their newly acquired genes. 1) Cells are first treated with a solution of CaCl2, which is thought to neutralize the repulsive negative charges of the phosphate backbone of DNA and the phospholipids of the cell membrane, allowing the DNA to adhere to the cells. This is referred to making the cells âcompetentâ (i.e. the cells are capable, or competent, to take-in exogenous DNA). 2) Cells are then subjected to a heat shock, which is thought to increase the permeability of the cell membrane, allowing the cells to take-in the plasmid DNA. 3) Cells are allowed to grow in a ârecoveryâ phase in the absence of ampicillin. During this time the cells begin to produce the Î²-lactamase protein, which will allow them to survive when they are subsequently placed on agar plates containing ampicillin.

For the first week section of lab

To eac

h of two microfuge tubes, add 250ul of transformation buffer. Keep these tubes on ice.

Using a sterile loop, pick one colony of bacteria from the LB plate. Transfer this colony to one of the microfuge tubes.

Gently vortex the tube until the colony is dispersed. Repeat for the second tube. Add 0.08ug of DNA to one of these tubes (+DNA tube).

Incubate both tubes on ice for 10 minutes. Incubate both tubes at 42Â°C for 50 seconds.

Incubate both tubes on ice for 2 minutes. Add 250ul of LB broth to each tube and incubate at room temperature for 10 minutes. Flick each tube gently. Add 100ul of t

his transformation mix to the appropriate plates.

Results For plates:

+DNA: Has the E. Coli pGlo plasmid and - DNA plates do no have EColi pGLo plasmid .

Amp means ampiclin and Ara means arabinose on agar plates

+ DNA Plate LB/amp/ara--> white 9 colonies, Glow: yes

+DNA LB/amp--> 7 colonies, GLow: no

--DNA LB/AMP- 1 colony so no colony growth and no glovw

-DNA: LB--> many colonies, Glow: no

Please explain significance of result or thand the reason for the results in scientific terms based on the background information provided and your own knowledge to be detailed. Would these results be expected as wll.

BIOL107 Lecture Notes - Lecture 3: Lipid Bilayer, Unsaturated Hydrocarbon, Alkane

BIOL107 Full Course Notes

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