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(1 pt each) If the following generally apply to all prokaryotes, write ‘P’. If they apply specifically to eukaryotes, write ‘E”. If they apply primarily to bacteria, write ‘B’, and if they apply primarily to archaea, write ‘A’. If it applies to more than one, include letters it applies to. 1 point each.

___P____6. DNA is organized in usually one large, circular chromosome.

____E___7. Cells contain membrane-bound organelles.

____P&B___8. Cell walls are common and contain some amount of peptidoglycan

____B&A___9. Small, circular pieces of DNA called plasmids, which are easily transferred, can often be found in cells.

____E___10. Ribosomes are found in cells and the entire ribosome size is ‘80S’, based on sedimentation rate.

____B&E___11. Phospholipids that comprise cell membrane feature fatty acids that are linked to glycerol by ester linkages.

__A_____12. Cell walls consist of protein/glycoprotein layer called S layer, pseudomurein layer, or in rare cases, only a slime layer outside of membrane(s).

___B&A____13. Typical shapes include bacilli and cocci.

____E___14. Cells have a cell membrane, DNA, and ribosomes.

_____A__15. Reproduction is asexual, and cells do not undergo mitosis or meiosis.

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Reid Wolff
Reid WolffLv2
28 Sep 2019

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Related questions

21) Which of the following is not primarily used for attachment?

A. glycocalyx

B. flagella

C. pili

D. fimbria

E. S layer

30) You have isolated DNA from an E. coli bacteria. The DNA would originate from what structures? (CIRCLE ALL THAT APPLY)

A. inner plasma membrane

B. mitochondria

C. microcompartments

D. nucleus

E. plasmids

31) Many cellular reactions require multiple enzymes. For efficiency, bacteria will package these enzymes together within:

A. fimbria

B. inclusion bodies

C. lysosomes

D. microcompartments

E. vesicles

32) Why can archaea be considered a separate domain from bacteria?

(CIRCLE ALL THAT APPLY)

A. Archaea ribosomal structure and protein synthesis are similar to eukaryotes.

B. They exhibit a novel method of DNA compaction.

C. Like eukaryotes, they contain simple, membrane-bound organelles.

D. Their cell walls are composed of different substances than bacterial or eukaryotic cell walls.

E. They contain a circular chromosome.

33) How are eukaryotic flagella different from bacterial flagella?

A. They are much bigger in size.

B. They move in a whip-like fashion, as opposed to a circular fashion.

C. They are covered by the plasma membrane.

D. A & B only

E. all of the above

36) The primary function of of the rough endoplasmic reticulum is:

A. aerobic respiration

B. the breakdown of phagocytized material

C. assembly of ribosomal subunits

D. polysaccharide and lipid modification of proteins

E. protein synthesis of membrane-bound and secreted proteins

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 thecell’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 → _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 cell’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 cell’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.


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 β-lactamase (bla), which provides resistance to the antibiotic ampicillin. The β-lactamase protein is produced and secreted by bacteria that contain the plasmid. β-Lactamase inactivates the ampicillin present in the LB nutrient agar, allowing bacterial growth. Only transformed bacteria that contain the plasmid and express β-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 cells’ 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 ‘competent’ (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 ‘recovery’ phase in the absence of ampicillin. During this time the cells begin to produce the β-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 42°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.

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