Leethaniel Brumfield, IIIGenetics – Spring 2007

GeneticsTable of ContentsLaboratory Experiments

Date Lab Title Pages

1-19-07 1: Laws of Probability & Chance(Chi-Square Test)

1 - 4

1-26-07 2: Mitosis(Onion Root Tip Experiment)

5 - 8

2-2-07 3: Mendelian Genetics 9 - 11

2-9-07 4: DNA Extraction from Plant Cells 12 – 16

2-16-07 5: Isolation & Purification of RNA(from Gram-Negative Bacterial Cells)

17 - 24

2-23-07 6: Separation of Unknown DNA Fragments Using Agarose Gel Electrophoresis

25 - 28

3-9-07 7: Applications of Recombinant DNA (Genome Analysis) & Genetics Online Lab

(DNA Databases)

29 - 33

3-16-07 8: Genetics of Human Taste Response 34 - 42

3-16-07 9: Drosophila Chromosomes 43 - 44

3-16-07 10: Human Barr Bodies 45 - 46

3-30-07 11: Whole Cell Protein Extraction(No Procedure)

47 - 49

4-13-07 12: Restriction Digestion of DNA 50 - 54

4-20-07 13: Transformation of E. coli(No Procedure)

55 - 60

4-27-07 14: Polymerase Chain Reaction Analysis 61 - 64

Lab #1: Laws of Probability & Chance

Background:1 toss = expectancy 50/50 probability (ratio) to get heads (H) or

tails (T)Independent events = occur simultaneously (at the same time)

df (Degree of Freedom) = N – 1* Law of Probability = probability of 2 or more independent

events occurring simultaneously is the product of their individual probabilities

(each event is multiplied together)* Chi-Square Test = used to determine if the experimental data (O) is a satisfactory approximation of the theoretical data (E)/ enables us to determine whether it is reasonable to attribute

deviations from a perfect fit to chance (probability).

Chart I:Tossing 1 Coin (1 Event – 100 Times)

Result (N)

Observed (O)

Expected(E)

O-E (O - E)2

(O–E)2/E X2 = ∑(O-E)2/E

df(N-1)

H 46 50 -4 16 0.32 1T 54 50 4 16 0.32 1

100(Total)

0.64

Probability = 0.50 > 0.30

Chart II:Tossing 2 Coins (2 Events Occurring Simultaneously – 96

Times)

Result (N)

Observed (O)

Expected(E)

O-E (O - E)2

(O–E)2/E X2 = ∑(O-E)2/E

df(N-1)

HH 21 24 -3 9 0.375 2

TT 20 24 -4 16 0.667 2

HT 55 48 7 49 1.020 2

96(Total)

2.062

Probability = 0.50 > 0.30

Chart III:Tossing 3 Coins (3 Events Occurring Simultaneously – 72

Times)

Result (N)

Observed (O)

Expected(E)

O-E (O - E)2

(O–E)2/E X2 = ∑(O-E)2/E

df(N-1)

HTH 26 27 -1 1 0.037 3

HHH 11 9 2 4 0.444 3

THT 25 27 -2 4 0.148 3

TTT 10 9 1 1 0.111 3

72(Total)

0.740

Probability = 0.90 > 0.70

Chart IV:* In a cross involving Drosophila (fruit flies) an F2 population included 272 flies w/ long, normal wings & 60 flies w/ dumpy

wings. Calculate Chi-Square using the results that approximately have a 3:1 ratio.

Expected Frequencies of Drosophila F2 Population 3:1 Ratio

Total = 332, NN = long, normal wings (272), nn = dumpy wings (60)

Result (N)

Observed (O)

Expected

O-E (O - E)2

(O–E)2/E X2 = ∑(O-E)2/E

df(N-1)

(E)NN 272 249 23 529 2.12 1

nn 60 83 -23 529 6.37 1

332(Total)

100% 8.49

Probability = 5.99 > 9.21

Chart V:* Some studies show that the following are the approximate

frequencies of the various ABO blood groups in the U.S. Hispanic population

Expected Frequencies of the Various ABO Blood Groups(Total = 839, Calculated 41% A, 9% B, 3% AB, and 47% O)

Result (N)

Observed (O)

Expected(E)

O-E (O - E)2

(O–E)2/E X2 = ∑(O-E)2/E

df(N-1)

A 326 (41%)343.9

-17.9

320.41

0.931 3

B 82 (9%)75.5

6.5 42.25 0.559 3

AB 27 (3%)25.1

1.9 3.61 0.143 3

O 404 (47%)394.3

9.7 94.09 0.238 3

839(Total)

100% 1.87

Probability = 0.70 > 0.50

Lab #2: Mitosis

Objective:To study mitosis in an onion root tip & be able to describe the

different mitotic stages.

Materials:an onion root, distilled H2O, test tube, tweezers, needles, Petri dish, acetocarmine, hot water bath, slides, cover slips, I M HCl,

alcohol lamp, microscope, thermometer, test tube rack

Procedure:See attachment.

Add to Step# 1: Take the root tip from the red stain & wash in distilled water (2x). After washing, place root tip in a test tube containing 1 M HCl. Keep test tube in hot water bath (for 10

mins.) at 60°C. Remove root tip from test tube & rinse in distilled water (3x).

Results & Conclusion:After processing the slides, Brittany Johnson & I were eventually able to see the different mitotic stages. We had to cut (tear) our root tips into very small pieces, which allowed us the ability to

observe specific mitotic stages in extremely identifying detail. The stages we observed were prophase & telophase. After

telophase was complete, we observed the mitotic cycle attempt to start again, beginning w/ interphase.

The main significance of mitosis is that the process stays the same. Haploid divides into haploid & diploid divides into

diploid. In meiosis, the process goes through reduction stages. For example, diploid divides into haploid.

Mitotic Stages:

1. (Interphase) DNA has replicated, but has not formed the condensed structure of chromosome. They remain as

loosely coiled chromatin. The nuclear membrane is still intact to protect the DNA molecules from undergoing

mutation.2. (Prophase) The DNA molecules progressively shorten

and condense by coiling, to form chromosomes. The nuclear membrane and nucleolus are no longer visible. The spindle apparatus has migrated to

opposite poles of the cell.

3. (Metaphase) The spindle fibers attach themselves to the centromeres of the chromosomes and align the

chromosomes at the equatorial plate.

4. (Anaphase) The spindle fibers shorten and the centromere splits, separated sister chromatids are pulled along behind

the centromeres.

5. (Telophase) The chromosomes reach the poles of their respective spindles. Nuclear envelope reforms before

the chromosomes uncoil. The spindle fibers disintegrate.

6. (Cytokinesis) This is the last stage of mitosis. It is the process of splitting the daughter cells apart. A furrow forms and the cell is pinched in two. Each daughter cell contains

the same number and same quality of chromosomes.

Note:Prophase & telophase were the 2 mitotic stages that Brittany Johnson & I

observed.

Lab #3: Mendelian Genetics

Background:Monohybrid cross = 1 trait/ 2 different kindsDihybrid cross = 2 traits/ 2 different kinds

P1 = parental generation

* There can be different genotypes for a specific phenotype.

Objective:To be able to find the parents, as well as the proceeding

generations using the Mendelian formula, & to also be able to find the genotype & phenotype ratios.

Materials:paper, pencil, eraser, sample parents, clear mind

Procedure:See attachment.

Results & Conclusion:See attachment.

Lab #4: DNA Extraction from Plant Cells

Background:Cellulose = main ingredient in plant cell wall

Peptidoglycan = man ingredient in bacteria cell wallChitin = main ingredient in fungal cell wall

* For DNA to be extracted (released), a thick cell wall (nuclear membrane), coated with a middle lamella, primary cell wall, & a secondary cell wall, must be penetrated to reach the cytoplasm., where many

organelles reside. During this process of DNA extraction, enzymes are used. This gets rid of the other unwanted

materials & components.

Objective:To extract DNA from wheat germs cells.

Materials:enzyme bottle, 100 mL beaker, distilled H2O (refrigerated), 95% alcohol, untreated wheat germs (refrigerated), mortar & pestle, lysing solution, hot water bath, Styrofoam box, ice. 4 sheets of

cheesecloth, Pipette, test tube, spooling rod, thermometer, timer

Procedure:See attachment.

Results:DNA fibers attached themselves to the glass rod. The fibers

resembled white mucous. It is anticipated that this extracted DNA, which was stored in the freezer, will be used later in an

electrophoresis experiment.

Discussion:The packet of wheat germs cells was grinded w/ a mortar &

pestle, in order to extract the DNA, since the cell walls of plants cells is much harder to lyse (split) than those of bacterial cells. A spooling rod was inserted into the DNA solution, along w/ alcohol (DNA is soluble in water but not in alcohol) – 95% (compared to 75%) alcohol enables a greater yield of spoiled DNA (as the rod

was rotated).The most stable molecule configuration for DNA is the

double helix. DNA is very lengthy, which makes it very susceptible for cleavage & extremely viscous in solutions. Base

pair sequences determine the amino acid sequence in the proteins made in the cell. The DNA fibers extracted from the solution were attached to the rod at the interface of 2 layers.

Conclusion:The DNA was extracted & coiled around the glass rods. The rods

were saved for future research.

Questions:1. What is the role of lysing solution which contains the

detergent?Answer: The role of the lysing solution is to break open the nuclear membrane, so the contained DNA can be released.

The lysing solution also emulsifies lipids and proteins, causing them to precipitate out of the solution.

2. What is the role of sodium chloride in the lysing solution?

Answer: Sodium chloride in the lysing solution reacts with the negative phosphate ends of the DNA, causing adjacent nucleic

acids molecules to coalesce (join together).

3. How are proteins denatured in this experiment?Answer: Though a temperature of 65°C was used, the proteins in this experiment were not denatured because doing so would

actually kill the DNA.

4. Describe the appearance of the extracted DNA.Answer: The extracted DNA resembled white, slimy mucous

that had been wrapped around the glass rod.

5. List the 4 kinds of nucleotides of DNA and explain what differentiates them from one another.

Answer: Adenine, guanine, cytosine, & thymine are the 4 kinds of nucleotides of DNA. The difference between the 4 has to do primarily with their nitrogenous bases. Adenine & guanine are

double-ring structures (purines), & cytosine & thymine are single-ring structures (pyridines).

6. What are the functions of DNA in a cell? Name the components of a bacterium, fungal, and plant cell

wall.Answer: The cells copy DNA to provide instructions for

constructing proteins & regulating their synthesis (transcription). Replications involves chain separation &

formation of complementary molecules of DNA on each free single chain, which attracts to itself the very sequences of nucleotides needed to rebuild itself. Each single chain then

serves as a template for the formation of its complement. This assures that each daughter molecule is identifiable to the

parent. Each daughter chain is composed of 1 strand from the parent & 1 newly synthesized complementary chain. The

components of a cell wall are cellulose (plant cell), peptidoglycan (bacteria), & chitin (fungal).

7. How many genes are present in a bacterium and how many base pairs are present in a typical gene?

Answer: A typical bacterium is composed of about 3,000 genes. A typical gene contains about 1,000 base pairs.

Lab #5: Isolation & Purification of RNA(from Gram-Negative Bacterial Cells)

Background:Gram-negative bacteria = are those that do not retain

crystal violet dye in the gram-staining protocol/ end color is the secondary (2nd) color

RNA is a nucleic acid polymer consisting of nucleotide monomers. RNA polynucleotides contain ribose sugars & predominantly uracil unlike

DNA, which contains deoxyribose &predominantly thymine. It is transcribed (synthesized) from DNA by enzymes called RNA

polymerases and further processed by other enzymes. RNA serves as the template for translation of genes into proteins, transferring amino

acids to the ribosome to form proteins, & also translating the transcript into proteins. Synthesis of RNA is usually catalyzed by an enzyme - RNA polymerase, using DNA as a template. Initiation of synthesis

begins with the binding of the enzyme to a promoter sequence in the DNA (usually found "upstream" of a gene). The DNA double helix is unwound by the helicase activity of the enzyme. The enzyme then

progresses along the template strand in the 3’ to 5’ direction, synthesizing a complementary RNA molecule w/ elongation occurring

in the 3’ to 5’ direction. The DNA sequence dictates where RNA synthesis will occur. There are also a number of RNA-dependent RNA polymerases as well that use RNA as their template for synthesis of a new strand of RNA. For instance, a number of RNA viruses (such as

poliovirus) use this type of enzyme to replicate their genetic material. Also, it known that RNA-dependent RNA polymerases are required for

the RNA interface pathway in many organisms.

* The cell wall of bacteria is made up of (gram-positive) peptideoglycan. A gram-negative bacterium is generally more resistant because the antibodies cannot penetrate

the lipopolysaccharide layer.

Objective:To isolate gram negative bacterial cells & purify RNA.

Materials:mircofuge tubes, centrifuge, 0.5 mL bacteria medium, ice bath, pipette, 300 µL lysis solution, incubator, tweezers, micropipette,

100 µL DNA precipitation, 300 µL 95% isopropyl alcohol, absorbent paper(paper towels), 300 µL 70% ethanol, 50 µL RNA

hydration solution

Procedure:See attachment for extensive, more thorough

specifications.

Briefly:I. Cell Lysis

1. centrifuge (10,000 RPM/ 1 min) collected bacteria sample tube from culture to pellet cells

2. put tube in ice bath3. decant supernatant & add cell lysis solution

4.incubate (@ 65°C/ 1 min) to lyse cells5. cool (@ room temp/4 mins)

II. Protein-DNA Precipitation1. add protein-DNA precipitation solution to cell lysate

2. put tube in ice bath3. centrifuge (10,000 RPM/6 mins)

the precipitated protein & DNA forms a light pellet at tube bottom

III. RNA Precipitation1. pour the RNA supernatant into another tube/leave

precipitate protein-DNA pellet2. add 95% isopropyl alcohol to supernatantDNA forms visible white clump in tube

3. centrifuge (10,000 RPM/3 mins) RNA forms pellet at tube bottom

4. pour off supernatant/ add 70% ethanol5. wash RNA pellet/ centrifuge (10,000 RPM/3 mins)

6. pour off supernatant/ leave RNA pellet in tube bottom/allow RNA pellet to dry (incubator/15 mins)

IV. RNA Hydration 1. add RNA hydration solution/ put RNA tube in ice bath (30

mins)RNA cells rehydrate

2. pipet well before use/ if not used immediately, store in freezer

Notable enzymes commonly used in molecular biology:

I. Ribonuclease A (RNase A)1. enzyme derived from bovine pancreas

2. used to remove RNA from DNA preparations by degrading RNA into oligoribonuleotides (too small to be alcohol

precipitated w/ DNA)

II. Ribonuclease T11. enzyme isolated from Aspergillus oryzae

2. used to remove RNA from DNA preparations by degrading RNA into oligoribonuleotides (too small to be alcohol precipitated w/ DNA)/ more effective when used

in conjunction w/ ribonuclease A

III. Nuclease S11. enzyme isolated from Aspergillums oryzae

2. used to remove single-stranded tails from DNA to produce blunt ends/ to remove unhybridized regions of

heteroduplex nucleic acids

IV. Deoxyribonuclease I1. bovine DNase I

2. degrade double & single-stranded DNA to oligo & mononucleotides

3. labeling by nick-translation, it generates nicks in double-stranded DNA/ removes DNA from RNA preparations/

produces random DNA fragment generation M13 clones to be used for DNA sequencing

V. Proteinase K1. serine protease w/ broad cleavage specificity isolated

from the fungus Tritirachium album Limber2. removes nuclease from RNA & DNA preparations/ used in

protein fingerprinting

Results & Conclusion:Isolation and purification of RNA was accomplished. RNA

hydration was successful. A RNA pellet was formed & dried, then stored for later use. However, it is only assumed that the isolated & purified pellet was RNA. Further investigation would be needed

for more reliable confirmation.

Questions:1. What are the different classes of RNA? Function?

Answer: The different classes are snRNA, mRNA, rRNA, & tRNA. snRNA is involved in maturation of transcription. mRNA is involved in transcripts of protein-coding genes. rRNA is

involved in the structural components of ribosomes. tRNA is involved in the facilitation attachment of amino acids by

ribosomes.

2. A human gene was initially identified as having 3 exons (456, 224, & 524) & 2 introns (2.3 kb & 4.6 kb). Draw the gene showing exons w/ interior expressed

gene.

See next page for drawing.

Lab #6: Separation of Unknown DNA Fragments Using Agarose Gel Electrophoresis

1 g = 100 mL = 1%

* This lab was actually separated into 2 parts:(1) Separation of DNA Fragments(2) Agarose Gel Electrophoresis

Background:(1)

ethidium bromide = light sensitive/ stains DNA (allows you to see the bands)

TBE = (buffer) Tris Borate Ethylene Diamine

(2)Agarose = natural polysaccharide powder used to make gel (from red algae/seaweed)/ the less used, the faster the DNA

will move through pores (less compact the gel is)/ low EEO (electroendosmosis –

causes band smear/broaden), which allows excellent band resolution/ agarose gel is held together by hydrogen bonds,

which can be broken heating (this makes them easy to create & pour/ is the matrix of choice for RNA & DNA separation

(large macromolecules)(Observation) DNA is loaded into the wells. The colored dye

is added to the colorless DNA (which has sugar that makes the DNA more dense). The phosphates are (-) charged & attracted

(migrate) to the (+) terminals. The lighter (less dense) DNA fragments move faster & separate completely.

* Electrophoresis = technique used to separate macromolecules (Ex: DNA, RNA, protein) from a mixture using electric current based on each macromolecule’s

electric charge & size.

Objective:(1) To separate DNA fragments.

(2) To prepare agarose gel for electrophoresis.

Materials:(1)

0.60g agarose, 40 mL TBE, heater, flask, tape, cylinder, marker, scissors, comb, mold plate, scoop, weighing paper, scale, glass

rod, electrophoresis apparatus, DNA sample (unknown & standard), microfuge tips, microfuge, power supply, ethidium

bromide, shaker, UV illuminator, H2O

(2)scale, heater, flask, spatula, comb, tap[e, side blocks, tray,

cylinder, microfuge tip, microfuge, red dye, 40 mL of TBE, 0.60 g of agarose (low EEO)

Procedure:(1)

Weigh 0.60 g of agarose. Mix with 40 mL of TBE buffer. Place on heater with solution. Start to boil, then let cool. Pour solution into gel mold plate. Let mold gel. Remove comb from gel & place into electrophoresis apparatus. Make sure to place the end with the wells on the negative side (block). Pour buffer into apparatus, filling it up enough to cover gel. Load a standard at 1 end. A

standard is a kb ladder, which is a known solution. Make sure that the unknown fragment pair. Each person will load 10 µL into a well. Keep DNA ladder free of contamination. Start the power

supply, setting it for 100 volts. Run for 45-60 mins. Cut corner of gel to identify gel. Put gel into ethidium bromide & place on

shaker for 4 mins. Place on UV illuminator. If dye is too dark, put

gel in H2O for 10 mins. Then place on UV illuminator again to record DNA markers.

(2)Measure buffer in cylinder & weigh agarose on scale then pour

into flask. Mix 40 mL of TBE with 0.60 g of agarose to equal 1.5% gel. Stir solution & place on heater until particles are dissolved. In

a clean tray with side blockers, pour solution & place comb in solution. Let solution stand for 15 mins. to completely mold. After

agarose gel has set, take the comb out of gel. Then load wells with 10 mL of dye. When finished, dispose of cast gel & clean

equipment used.

Results & Conclusion:(1)

The gel set & was loaded properly. Since the well was illuminated correctly, the markers could be recorded.

See drawing below for additional results.(Ex: size of unknown fragment.)

(2)As expected, the structure gelled after 15 mins. The wells

were formed & dye was loaded into wells. It remained & was completely confined. I concluded that if the agarose gel was a

real DNA solution I would have been able to splice it & study the DNA.

Lab #7: Applications of Recombinant DNA (Genome Analysis) & Genetics Online Lab (DNA

Databases)

* This lab was actually separated into 2 parts:(1) Applications of Recombinant DNA (Genome Analysis)

(2) Genetics Online Lab (DNA Databases)

Background:(2)

Annotation = identifying genes in DNA sequenceNCBI = National Center for Biotechnology InformationHuman Genome Resources = OMIM (Online Mendelian

Inheritance in Man)Cytochrome P450 = a protein

Entry = the life science search engineGetting into databases = Pubmed (NCBI- Molecular Database)

BLAST = Base Local Alignment Search ToolBlastp = protein-protein BLAST

Nucleotide, protein, translation (retrieve results for an RID)Database = Expasy (Expert Protein Analysts System)

Database = Swiss-Prat-TrEmble Knowledge Base

Objective:(1)

To apply specified applications of recombinant DNA genome analysis.

(2)To access the provided genetic online database

(http://www.dnaftb.org/dnaftb/) to perform and complete a series of genetic questions.

Materials:(1)

lecture textbook, pencil, paper, clear mind(2)

online capabilities, useful genetics website references, pencil, paper, clear mind

Procedure:(1)

See attachment (Specified questions were completed).(2)

Go to the following website: http://www.dnaftb.org/dnaftb/)Click on “Molecules of Genetics”…

1. Click on Classic Genetics: #15 (DNA & Proteins Are key Molecules of the Cell Nucleus).

Click on “Problem”(Read and answer the series on multiple choice questions)2. Click on Classic Genetics: #10 (Chromosomes Carry

Genes). Click on “Problem”

(Read and answer the series on multiple choice questions)3. Click on Classic Genetics: #19 (The DNA Molecule is

Shaped like a Twisted Ladder). Click on “Problem”

(Read and answer the series on multiple choice questions)

4. Click on Classic Genetics: #39 (A Genome is an Entire Set of Genes). Click on “Problem”

(Read and answer the series on multiple choice questions)5. Click on Classic Genetics: #40 (Living Things Share

Common Genes). Click on “Problem”(Read and answer the series on multiple choice questions)

Results & Conclusion:The following were completed successively:

(1) Applications of Recombinant DNA (Genome Analysis)(2) Genetics Online Lab (DNA Databases)

Refer to “Problems” for an extended analysis of results.

Problems:(1)

1. Question 8.12 – (See attached handout)Answer: The Scott’s aren’t the parents. The Larson’s are the parents, since

James’ sequence has similar fragments. There are several fragments in James’ sequence that aren’t present in the Scott’s.

2. Question 8.13 – (See attached handout)Answer: T cells were isolated & grown in the lab and normal DAD was

introduced using a viral vector.

3. Question 8.14 – (See attached handout)Answer: Insulin may not have been expressed due to it not being inverted

into the reading frame, which would disable effective processing.

4. Question 8.15 – (See attached handout)Answer: In my opinion, the concern isn’t justified because altering a plant

presents various effects that have the ability to result in different phenotypic expressions.

(2)1. Question 9.16 – (See attached handout)

Answer: Annotation, the next step after a genome is completely sequenced, is the actual identifying of genes in DNA sequence. During annotation,

putative genes & other important sequences are identified & described by:a. describing the functions of all genes of the organism (protein-coding

should be of particular interest)b. involves using computer algorithms (to search both DNA strands &

sequences for protein-coding genes)c. putative-protein-coding genes arte found by using ORFs (start codons in

frame with a stop codon)* Annotation = # of genes, size of genome, # of proteins, length of

chromosomes

2. Question 9.17 – (See attached handout)

Answer: DNA microarrays can be, indeed, useful in functional genome approach to understand human diseases that have environmental

components, such as cancers, since functional genomics concentrate on gene expression (control), physiology, & development. DNA microarrays, already

considered a valuable tool for studying the transcriptome, could be incorporated into bioinformatics research to assess how disease phenotypes arise from the interaction of genes with their environments, to describe the

interactions between genes & genes products at a global level within the cell, between cells, and between organisms, and to postulate phylogenic

relationships for sequences.

3. Question 9.19 – (See attached handout)Answer:

a. S, F, C = Aligning DNA sequence within databases to determine the degree of matching

b. F, C = Annotation of sequences within a sequenced genomec. F = Characterizing the transcriptome and proteome present in a cell at a

specific developmental stage or in a particular disease stated. C = Comparing the overall arrangements of genes and nongene sequences in different organisms to understand how genomes evolve

e. F = Describing the function of all genes in a genomef. F, C = Determining the function of human genes by studying their

homologs in nonhuman organismsg. F = Developing a comprehensive two-dimensional polysaccharide gel

electrophoresis map of all proteins in a cellh. S = Developing a physical map of the genome

i. S, F, C = Developing DNA microarrays (DNA chips)j. F, C = Identifying homologs to human disease gene in organism suitable

for experimentationk. S = Identifying a large collection of simple tandem repeat or microsatellite

sequences to use as DNA markers within one organisml. S, F, C = Identifying expressed sequence tags

m. F, C = Making gene knockouts and observing the phenotypic changes associated with them

n. S = Mapping a gene in one organism using the lod score methodo. S = Sequencing individual BAC or PAC clones aligned in a contig using a

shotgun approachp. S = Using oligonucleotide hybridization analysis to type an SNP

4. Question 9.20 – (See attached handout)a. Mycoplasma genitalium has a 580 kb genome sequence that encodes for

480 proteins.b. The genes required for life are those that are consistently present in

organisms.c. Yes, there could be genes of unknown function that are essential to life.  If the genes are knocked out in a model animal we could test for differences in gene and protein expression in the animal to examine potential functions of

that gene.d. The following would be helpful to discern the functions of protein-coding

genes: (1) aligning DNA sequences within databases to determine the

degree of matching, (2) annotating sequences within a sequenced genome, (3) characterizing the trascriptome and proteome present in a cell at a specific developmental stage or in a particular disease, determining the

functions of human genes by studying their homologs in nonhuman organisms, (4) developing a comprehensive two-dimensional polyacrylamide

gel electrophoresis map of all proteins in a cell, (5) developing DNA microarrays, (6) identification of homologs to human disease genes in

organisms suitable for experimentation, (7) identifying expressed sequence tags (ETS), (8) making gene knockouts, and (9) observing the phenotypic

changes associated with them.e. The definition of life is a widely debated subject. However, identifying a

minimal genome would aid in answering the fundamental question in biology, “What is life?”

f. Some sequences that are unable to be synthesized may have segments essential for life, but if all of the DNA sequences were able to be synthesized it would be possible to assemble these sequences into a chromosome. It is a

possibility that the sequence might code for different things in different organisms. For example, in Mycoplasma genitalium, UGA codes for

tryptophan, yet in E. coli it codes for a stop codon, leading to premature termination.  Additional information would be needed for E. coli to know that

the UGA codes for tryptophan and not for a stop codon.g. The organism in (f) would have early termination when compared to

Mycoplasma genitalium.h. These analyses of genes could stem from the synthesis of entire

organisms from these genes. For quite sometime, cloning is an ethical issue in contemporary society. Scientists in collaboration with the government

(complete with its checks and balances) should decide how to address these issues.

5. Question 9.21 – (See attached handout)Answer:

a. The genes on chromosomes V & X are distributed uniformly, but chromosome V has conserved genes found more frequently in the central regions, while inverted and tandem-repeat sequences are

found more frequently on the arms. Chromosome V appears to have a more inverse relationship between frequency of inverted & tandem-

repeats & frequency of conserved genes.b. Because there are less conserved genes on the arms, there seems to

be a greater rate change on chromosome arms when compared to the central regions.

c. Indeed. Because increased meiotic recombination results in greater rates of change of genetic material on chromosome arms.

Lab #8: Genetics of Human Taste Response

Background:Taste Test Paper (PTC):

1. Arthur Fox: 1st to use compounds (like PTC/phenylthiourea), which are bitter to some people & tasteless to others & has been an important character in population genetics investigation & racial population

2. PTC: ability to taste PTC is inherited/ the taste dimorphism attaches to a pair of genes

3. people with homozygous recessive genes are nontasters, which is broken down into different races as: North American White = 0.550 (20% from tasters, 50% nontasters), Japanese = 0.266, Jews = 0.524, Hindu = 0.581, & Indian (Brazil) = 0.111

4. to be tested, PTC has to be dissolved in the taster’s own saliva/ if PTC is dissolved in another’s saliva or in H2O & placed on a dry tongue of taster, it won’t be tasted.

5. some people have a higher taste sensitivity threshold for PTC than others (Blakeslee)

6. classification into tasters & nontasters is just a reflection of the bimodal distribution of taste threshold

7. sodium benzoate (similar to PTC) use is limited because some believe that its use is detrimental to health

8. Fox categorized tasters & nontasters for PTC into 5 subgroups, depending on whether sodium benzoate is present – salty, sweet, sour, bitter, tasteless

Taste Modalities (5)Genetics of Human Taste Response

I. Taste Modalities1. Umami – savory flavor exemplified by glutamate

(amino acid)2. 5 subgroups (mediated by taste receptor proteins

on taste receptor cells (TRCs) within the taste buds on tongue

3. TRCs = send signal transduction pathways to indicate taste depolarizing by interacting w/ ion channel

4. Amino acids, sugars, etc = perceived as sweet & bitter activate G protein-coupled receptors (GPCRs)

5. Very little is known about:a. Variation in these taste sensesb. Heritability of this variation

II. Bitter1. Most complex taste quality in humans

a. Wide variety of chemical structures that elicit bitterness

b. Large # of genes encoding receptors for the taste modality

2. Bitter taste receptor genes (in humans):a. T2R (24 potentially functional)b. TAS2R (several pseudogenes)

3. The bitter taste is perceived by many to be unpleasant, sharp, or disagreeable. Evolutionary biologists have suggested that distaste for bitter substances may have evolved as a defense mechanism against accidental poisoning. However, not all bitter substances are harmful; coffee and tonic water are both popular, bitter beverages. Other common bitter foods include

bitter melon, arugula, watercress, and dandelion. Quinine, the anti-malarial prophylactic, is also known for its bitter taste and is found in tonic water. Research has shown that TAS2Rs (taste receptors, type 2) coupled to the G protein gustducin are responsible for the human ability to taste bitter substances. They are identified not only by their ability to taste for certain "bitter" ligands, but also by the morphology of the receptor itself (surface bound, monomeric).

III. PTC (phenylthiourea)1. The ability for a person to taste PTC (&

structurally related compounds) is the best studied example of a person’s taste response to bitter compounds (tasters & nontasters)

2. PTC sensitivity is majority effected by a gene on chromosome 7 & another on chromosome 16

3. The gene on chromosome7 is in the TAS2R bitter taste receptor gene (TAS2R38)

IV. Sweet & Umami1. Both indicate the presence of calorically rich &

essential nutrients2. Very different perceptually, but similar

phylogentically3. Umami taste – performed with (MSG) monosodium

glutamate4. Sweetness is produced by the presence of sugars,

some proteins and a few other substances. Sweetness is often connected to aldehydes and ketones which contain carbonyl group. The compounds which the brain senses as sweet are thus compounds that can bind with varying bond strength to several different sweetness receptors. The differences between the different sweetness

receptors are mainly in the binding site of the G protein coupled receptors.

5. Umami is the name for the taste sensation produced by the free glutamates commonly found in fermented and aged foods. Examples of food containing these free glutamates (and thus strong in the savoury taste) are parmesan and roquefort cheese as well as soy sauce and fish sauce. It is also found in significant amounts in various unfermented foods such as walnuts, grapes, broccoli, tomatoes, and mushrooms, and to a lesser degree in meat. The glutamate taste sensation is most intense in combination with sodium. This is one reason why tomatoes exhibit a stronger taste after adding salt.

6. The additive monosodium glutamate (MSG) produces a strong savoury taste.

7. A subset of umami taste buds responds specifically to glutamate in the same way that sweet ones respond to sugar. Glutamate binds to a variant of G protein coupled glutamate receptors.

V. Sour1. A basic taste quality that ignites an innate

rejection response2. Acidic stimuli are the unique sources of sour taste3. Sourness is the taste that detects acids. The

mechanism for detecting sour taste is similar to that which detects salt taste. Hydrogen ion channels detect the concentration of hydronium ions (H3O+ ions) that have dissociated from an acid.

4. Hydrogen ions are capable of permeating the amiloride-sensitive sodium channels, but this is not the only mechanism involved in detecting the quality of sourness.

VI. Salty1. Guides consumption of NaCl (serving an essential

function in ion & H2O homeostasis)2. A sodium channel sensitive to the channel-blocker

amiloride serves as a salt-taste receptor by providing a specific pathway for sodium current into the taste cell

3. The amiloride sensitivity of NaCl taste in humans is specific to the very minor sour component & not to the salty taste itself.

4. The human salty taste response to NaCl be blocked by the addition of the compound chlorhexidine.

5. Saltiness is a taste produced by the presence of sodium chloride (and to a lesser

degree other salts). The ions of salt, especially sodium (Na+), can pass directly

through ion channels in the tongue, leading to an action potential.

Objective:To be able to track and trace random tasters’ taste response and family pedigree.

Materials:random tasters (at least 3), taste strips (control, phenylthiourea/PTC, sodium benzoate, thiourea), sterile Petri dish and lid, family pedigree forms, pencil, paper, clear mind

Procedure:Obtain 3 of each taste strip type. Place them in a sterile Petri dish. Seek random tasters. Record results in a chart-type diagram to enable taste response analysis. Ask the necessary question to obtain a valid family pedigree diagram.

Results:

See diagram below.

Human Type

Control - White

Phenylthiourea

(PTC) - Blue

SodiumBenzoate

- Pink

Thiourea - Yellow

Bittersweet(Sour, Salty)

Taster

Non-Taster

Taster #1

Female (1st)

Taster #2

Female (2nd)

Taster #3

Male

*Represents positive taste response =

Note:1st Female = same food, different drinks (grandparent,

parents/siblings)2nd Female = same food, same drinks (parents & siblings)

Male = same food, different drinks (parent/siblings)“How Taste Response Is Hard-wired Into the Brain”

Source:

Cell Press

Date: January 25, 2006

More on:

Neuroscience, Brain Injury, Perception, Nutrition Research, Disorders and Syndromes, Behavior

Science Daily — Instantly reacting to the sweet lure of chocolate or the bitter taste of strychnine would seem to demand that such behavioral responses be so innate as to be hard-wired into the brain. Indeed, in studies with the easily manipulable fruit fly Drosophila, Kristin Scott and colleagues reported in the January 19, 2006, issue of Neuron experiments demonstrating just such a hard-wired circuitry.

Their findings, they said, favor a model for taste encoding in the brain that holds that specific cells are dedicated to detecting specific tastes. Competing models hold that multiple neurons combine information to encode taste, or that the timing of patterns of taste information encodes taste.

In their studies, the researchers explored the behavioral effects of activating fly taste neurons that had either of two chemical taste receptors on their surface. In earlier studies, the researchers had shown that the Gr5a receptor on taste neurons was essential for response to sugar and that the Gr66a receptor was essential for response to bitter tastes. However, those studies left open the question of whether those different neurons selectively detected the different tastes and whether they generated taste behaviors.

To directly monitor taste responses of the flies, the researchers generated flies with fluorescent labels on their neurons that would signal activation of one or the other type. They used microscopic imaging through tiny windows in the fly brains to watch neuronal response when they exposed the flies to sweet or bitter chemicals

They found that a whole range of sweet substances selectively switched-on the Gr5a neurons, while a range of bitter substances switched-on the Gr66a neurons. However, the "sweet neurons" did not respond to bitter substances, and vice versa.

In behavioral studies, they found that flies preferred to spend time tasting substances that activated the Gr5a neurons and avoided substances that activated Gr66a neurons.

In the most telling experiments, the researchers engineered flies so that the hot pepper compound capsaicin would selectively switch-on either the sweet-detecting Gr5a neurons or the bitter-detecting Gr66a neurons. Normal flies do not respond to the hot pepper taste.

The researchers found that flies engineered to recognize capsaicin on the sweet-tasting neurons were attracted to the chemical, while those that recognized it as a bitter taste avoided it.

"In this paper, we demonstrate that these taste cells selectively recognize different taste modalities, such that there is functional segregation of taste qualities in the periphery and at the first relay in the brain," concluded the researchers. "Moreover, we show that activation of these different taste neurons is sufficient to elicit different taste behaviors. Thus, activity of the sensory neuron, rather than the receptor, is the arbiter of taste behavior.

Our studies argue that animals distinguish different tastes by activation of dedicated neural circuits that dictate behavioral outputs," they wrote.

Lab #9: Drosophila Genetics

Objective:To be able to thoroughly understand polytene’s association with

the genetics of Drosophila.

Materials:

lecture textbook, pencil, paper, clear mind

Procedure:Obtain handouts from professor (Dr. Abraham) and reference the

lecture textbook in order to understand polytene’s association with the makeup of Drosophila. Use the drawings and charts,

along with the copy in the lecture textbook to answer the questions specified by the professor.

Results & Conclusion:In all questions were answered with assured confidence.

Questions:1. What are polytene? Where are they found?Answer: Polytene chromosomes are special kinds of

chromosomes found in tissues (such as the salivary glands in the larval stages) of insects of the order Diptera (Drosophila). In polytene chromosomes, the homologous chromosomes are

tightly paired. Therefore the observed number of polytene chromosomes per cell is reduced to half the diploid number of chromosomes. Polytene chromosomes can be 1,000x the size of corresponding chromosomes at meiosis or in the nuclei of ordinary somatic cells and are easily detectable under the

microscope.

2. How many pairs of chromosomes does Drosophila have? What are the sex chromosomes for male and

female?Answer: Drosophila has 4 pairs of chromosomes. The sex chromosomes are XX (female) and XY males, which is the

same for humans.

3. Draw a figure of Drosophila polytene chromosome and tag the parts.

Answer: See next page for drawing.

Lab #10: Human Barr Bodies

Background:In those species in which sex is determined by the presence of the Y or

W chromosome rather than the diploidy of the X or Z, a Barr body is the inactive X chromosome in a female cell, or the inactive Z in a male, rendered inactive in a process called lyonization. The Lyon hypothesis

states that in cells with multiple X chromosomes, all but one is inactivated during mammalian embryogenesis. Barr bodies are named

after their discoverer, Murray Barr.

Objective:To be able to detect Barr bodies in a human female cell.

Materials:lecture textbook, pencil, paper, clear mind, a sample human

female cell slide, microscope (preferably binocular)

Procedure:Obtain a sample human female cell slide and view it under a

microscope (binocular in the effort to detect Barr bodies. Reference lecture textbook to gather a more thorough

understanding of Barr bodies and their significance to genetics. Incorporate this knowledge in answering the professor’s (Dr.

Abraham) questions.

Results:See sketch below.

(For sketch of Barr body that was detected under the microscope from the slide.)

Questions:

1. What is a Barr body? Where are they found in humans?Answer: A Barr body is a highly condensed & highly transcriptionally

inactive X chromosome found in the nuclei of somatic cells of female mammals.

2. How many Barr bodies does a Klinefelter Syndrome male have?

Answer: 1-2 Barr bodies, depending on the total number of chromosomes he has.

3. How is the number of Barr bodies in an individual related to the X chromosome?

Answer: Barr bodies are associated with the X chromosomes because their ability to form is dependent on the X chromosomes

being inactive.

Lab #11: Whole Cell Protein Extraction

Background:lysozyme = splits (breaks) cell wall to get proteins

centrifuge = (centrifugal force/spin) used to separate supernatant (unwanted) from bacteria cells./proteins (intact pellet)

rotor = center piece of centrifuge machinehemoglobin = protein enzyme used to carry oxygen through

blood

Objective:To extract and isolate whole cell proteins of a sample bacteria

culture.

Materials:pipette (100 µL-1000 µL), tweezers, microfuge tubes, 500 µL of

whole cell bacteria, gloves, centrifuge machine, micro tips (1000 µL & 10 µL), marker (for labeling), distilled H2O (500 µL & 450 µL),

beaker, 1.5 µL lysozyme, incubator, chloroform (150 µL), methanol (600 µL & 300 µL), vortex machine, boiling H2O

Procedure:In brief:

1. spin (centrifuge) 500 µL cells (12 mins @ 6000 RPMs @ 7°C

2. remove top layer & save pellet3. add 500 µL of distilled H2O (sterile)4. break up pellet (dissolve)5. add 1.5 µL of lysozyme & incubate (45 mins @

37°C)6. boil (10 mins)7. spin (1 min) to collect debris in bottom8. keep supernatant & transfer to a new tube9. add 600 µL methanol, 150 µL chloroform, & 450

µL of distilled H2O10. vortex (1 min)/ spin @ 6000 RPMs (5 mins)11. discard upper layer, leaving interface &

lower layer12. add 300 mL of methanol & invert (4x)

13. spin @ 6000 RPMs (6 mins)14. remove supernatant & air dry pellet (10

mins) to remove residual methanol

Results & Conclusion:The centrifuge technique was successful, and the supernatant

was extracted. An RNA pellet was formed and observed (visible). To examine the RNA pellet, an electrophoresis gel would need to

be run. This is possible because the RNA pellet was saved for future research (use).

Questions:1. What are proteins (exactly) structurally speaking?

List some functions of proteins. Answer: Proteins are macromolecules made of 1 or more polypeptides. The functions of proteins depends on their

complex folded shape & composition, which can be classified as either primary, secondary, tertiary, & quaternary. Some of

the functions of proteins are the following:A. Antibodies - defend the body from germs

B. Contractile Proteins - responsible for movementC. Enzymes - speed up chemical reactions/Example:

Hemoglobin = carries oxygen through the bloodD. Storage Proteins - store amino acids

2. In each of the following cases, stating how a certain protein is treated, indicate how the level of the protein structure will change as a result of the

treatment:A. hemoglobin stored in a hot incubator @ 80°C

Answer: The hemoglobin protein will dissociate into its 4 subunits because heat destabilizes the ionic bonds of its

quaternary structure.

B. egg white (albumin) is boiledAnswer: The albumin protein will denature when its tertiary

structure is destabilized by heating (boiling), not allowing itself to retain the folded pattern need for solubility.

C. RNase is heated to 100°CAnswer: The RNase protein will denature when its tertiary

structure is destabilized but unlike albumin, RNase re-nature if cooled slowly, reestablishing its normal, functional tertiary

structure.

D. meat in your stomach is digested (gastric juices contain proteolytic enzymes)

Answer: The meat proteins will denature, most likely, when its tertiary & quaternary structures are destabilized by acidic

conditions in the stomach, and then the primary structure of the polypeptides will be destroyed as they degrade into their

amino acid components by proteolytic enzymes in the digestive tract.

3. If codons were 4 long, how many codons would exist in a genetic code?

Answer: The # of nucleotides raised to the power of the # of bases, which would be 44 = 256, would be the total # of

codons in the genetic code.

4. Contrast the component of prokaryotic ribosomes with eukaryotic riboosmes.

Answer: Both prokaryotic and eukaryotic ribosomes consist of 2 equally sized subunits, with complexity of at least 1 rRNA molecule & many ribosomal proteins. However, eukaryotic

ribosomes vary in composition & are larger & more complex.

Lab #12: Restriction Digestion of DNA

Background:Key points:

(1) RESTRICTION ENZYMES: (restriction endonucleases) (i) produced by various strains of bacteria & they recognize specific sequences in DNA molecules & then cleaves the phodiester bonds between the nucleotides(ii) type II = (most widely used) recognize & cut within specific DNA sequences, resulting in restriction fragments (DNA fragments)/ most type II make cuts 1 or 2 bases from axis of symmetry, resulting in the formation of single-stranded ends 2 or 4 bps long on the end of each fragment (iii) digestion of different DNA molecules w/ the same enzyme will yield a different set of restriction fragments, allowing such restriction analyses to be used as a fingerprint to characterize DNA molecules(iv) recognizes palindromic (symmetric) sequences in DNA (Example: EcoR I, considered a palindrome because the complementary strand has identical sequences in the opposite direction)(v) although restriction enzymes may be cut anywhere, a given enzyme always cuts between same 2 nucleotides/some cut in the center of sequence, resulting in the formation of “blunt ends” after both strands are cut - these complementary ends are also known as “cohesive” or “sticky ends”(vi) if both are cut w/ same enzyme, restriction fragments from other sources, which allows the formation of recombinant DNA molecules is the basis of most DNA cloning protocols (Note: cuts = restriction site/ cleavage)(vii) some restriction enzymes result in “blunt ends” that do not stick together but are useful in cloning because they can be attached to other “blunt ends”

(2) LAMBDA BACTERIOPHAGE:(virus that infects bacteria, specifically E. coli)

(i) linear double-stranded DNA molecule that has 12 base single-stranded complementary ends that allows lambda to anneal to form a circular molecule after it infects an E. coli cell(ii) circular molecule is 48,502 bp (48.502 kb) in length(iii) has been modified to develop a variety of cloning vectors

(3)DYE:(i) excess dye is not harmful(ii) it contains a sugar that holds (weighs) down its presence in an agarose electrophoresis solution, enabling it to be confined in the wells for identification

Objective:To study restriction digestion using lambda bacteriophage DNA

with restriction enzyme Bgl II.

Materials:lambda bacteriophage, Bgl II restriction enzyme, restriction

buffer, H2O (sterile), tweezers, centrifuge, microfuge, microfuge tips, microtube, marker, incubator

Procedure:(1) restrict digestion enzyme, (2) agrose gel electrophoresis,

(3) stain gel & study fragments

Obtain lambda bacteriophage DNA. Pipet 1 µL of lambda phage DNA in microfuge tube. Take 7 µL of sterile water, put in

microfuge tube. Add 1 µL of restriction buffer, then add 1 µL of restriction enzyme (Bgl II), which will total 10 µL of solution.

Thump microfuge tube a few times, and place in centrifuge for 5-7 secs. Put microfuge tube in float, and place in a water bath at

37°C for about 60 mins. After incubation, remove microfuge tube from float and perform agarose gel electrophoresis. Finally, stain

gel and study fragments.

Results & Conclusion:Restriction digestion, using lambda bacteriophage DNA, with

restriction enzyme Bgl II, was performed. However, the outcome was not completely reliable. There was slight error in the

procedure but we were not able to detect the bands (fluorescence). The intended 6 Bgl II fragments were not

identified or calculated, and a standard was not loaded into a well during the gel electrophoresis, which was vitally important during

the procedure because it would have verified the size of all identified fragments and confirmed or denied overall success.

Agarose Gel ElectrophoresisAgarose gel electrophoresis separates DNA fragments according to their size. Typically, a DNA molecule is digested with restriction enzymes, and the agarose gel electrophoresis is used as a diagnostic tool to visualize the fragments. An electric current is used to move the DNA molecules across an agarose gel, which is a polysaccharide matrix that functions as a sort of sieve to help "catch" the molecules as they are transported by the electric current. This technique has lots of applications. Generally speaking you can analyze DNA fragments that result from an enzyme digestion of a larger piece of DNA to visualize the fragments and determine the sizes of the fragments. In addition to its usefulness in research techniques, agarose gel electrophoresis is a common forensic technique and is used in DNA fingerprinting.

The phosphate molecules that make up the backbone of DNA molecules have a high negative charge. When DNA is placed on a field with an electric current, these negatively charged DNA molecules migrate toward

Couldn't you just dye?

Ethidium bromide is an intercalating dye, which means it inserts itself between the bases that are stacked in the center of the DNA helix. One ethidium bromide molecule binds to one base. As each dye molecule binds to the bases the helix is unwound to accommodate the strain from the dye.

Closed circular DNA is constrained and cannot withstand as much twisting strain as can linear DNA, so circular DNA cannot bind as much dye as can linear DNA.

.

the positive end of the field, which in this case is an agarose gel immersed in a buffer bath. The agarose gel is a cross-linked matrix that is somewhat like a three-dimensional mesh or screen. The DNA molecules are pulled to the positive end by the current, but they encounter resistance from this agarose mesh. The smaller molecules are able to navigate the mesh faster than the larger one, so they make it further down the gel than the larger molecules. This is how agarose electrophoresis separates different DNA molecules according to their size. The gel is stained with ethidium bromide so you can visualize how these DNA molecules resolved into bands along the gel. Southern blotting may also be used as a visualization technique for agarose gels. Unknown DNA samples are typically run on the same gel with a "ladder." A ladder is a sample of DNA where the sizes of the bands are known. So after you run out your sample, you can compare the unknown fragments to the ladder fragments and determine the approximate size of the unknown DNA bands by how they match up to the known bands of the ladder.

Lab #13: Transformation of E. coli

Background:Genetic Exchange = (3 types) transformation, transduction,

& conjugation

Transformation = DNA is released from cells into the surrounding medium & recipient cells incorporate it into

themselves from this solution/ was the 1st bacterial genetic exchange mechanism (1928)

Competent cells = cells that are in a state of being capable of undergoing natural transformation/ many bacteria cells gain

“competence” depending on certain environmental conditions & many other bacteria can be made competent by exposure to

artificial treatment (like E. coli)Note: E. coli can be made competent by exposing its cells to CaCl2. These newly competent cells are then receptive

to an insertion of foreign DNA contained in a plasmid.

Plasmids = (many bacteria contain at least 1 plasmid) are circular molecules of double-stranded DNA that function in many

aspects as small chromosomes but unlike chromosomes, they are dispensable (not essential, unimportant)/ they are self-replicating

& capable of encoding a variety of cellular functions/ none are known to encode essential cellular functions/ there are several

dozen plasmids in 1 cell/ many plasmids, called R factors, carry genes that that are resistant to antibiotics on the host cell/ plasmid genes often encode enzymes that that chemically

inactivate the drug &/or (by active transport) eliminate it from the cell

GFP gene = (Green Fluorescent Protein) has an aquatic origin – from a bioluminescent jellyfish that emits a green glow/ the

biological significance of this tumescence is unknown/ GFP glows by itself (autoflorescent in the presence of ultraviolet light), which

makes it widely used in research as a reporter molecule (a protein (like GFP) linked to a “mystery” protein, which can be

studied according to its location with the reporter molecule (GFP))

pGREEN plasmid = contains a GFP gene & a gene for ampicillin resistance/ has a mutant version of the GFP that turns bacteria “yellow-green”, even in normal light & florescent when exposed

to UV.“glowing” = (pGREEN plasmid) indicates successful

transformation in the E. coli plasmid (vector)

Objective:To investigate 1 mechanism of (artificial) genetic exchange

through the insertion of pGREEN DNA plasmid, which carries the gene for antibiotic resistance to ampicillin, into competent cells.

Materials:tubes (sterile capsules & lids), CaCl2 (250 µL cold), ice bath,

incubator, pGREEN DNA plasmid (10 µL), plates (Petri), hot H2O bath (heat shock), luria broth (250 µL), micropipet, ampicillin, E.

coli, burners (sterilization)

Procedure:See attachment.

Note: My partner was Devon Daniels.

Leethaniel’s Plate: - Plasmid LBDevon’s Plate: + Plasmid LB/ Ampicillin

Results:Unfortunately, Devon and I were unable to observe

transformation. Insertion of pGREEN DNA plasmid into competent cells, with regards to Devon’s plate (+Plasmid LB/ Amplicillin) was not successful. We were unable to compare

and contrast transformation of E. coli, and we were also unable to observe the significance of adding pGREEN plasmid DNA to

E. coli.

Lab #14: Polymerase Chain Reaction Analysis

Background:thermal cycler = automatically cycled reaction through

programmed temperature changes

PCR:(1) discovered in 1983 – Kay Mullis

(Chemistry Nobel Prize 1993)(2) has the ability to amplify (makes

copies of) specific sequences of DNA

complex mixtures w/o having to clone the sequence in a host organism

(3) when repeated, there is a geometric increase in the amount of unit-length DNA

(4) an important tool in modern molecular biology/a type of genetic engineering technique/ more sensitive & quicker than cloning

(5) uses primers that complement the base pairing of the specific DNA fragment being amplified to isolate a single –stranded sequence from a total DNA or RNA genome (no library needed)

(6) critical PCR parameters:(a) incubation time(b) incubation temperature(c)concentration of:

(1) Taq DNA polymerase(2) primers(3) MgCl2(4) template DNA

(7) when successful, allows DNA fingerprinting

(8) successful PCR results in an overall # of bps that is the same (exactly) as the primers used

(9) gives 1 product of a specific size that can be separated from the unused primers & dNTPs – “primer-dimer” is commonly produced (primers that can hybridize at the 3’ ends), which results in a low molecular weight at least 90% identical to reaction product

(10) produces “mis-priming” when a primer target band is mismatched, which only affects 1st amplification cycle. Using the highest annealing temperature possible minimizies “mis-priming”(11) 50 – 100 µL (maximum total band solution amount)

Objective:To use PCR to obtain a PCR product from a DNA fragment.

Materials:thermal cycler, centrifuge, microcentrifuge tubes, micropipette tips, micropipette, autoclaved distilled H2O (78.5 µL), PCR buffer w/ MgCl2

(16 µL), 10 mM dNTP mixture (2 µL), 2 primer mix – primer I: 9877, primer II: 9899 (2 µL), control DNA (1 µL), Taq DNA polymerase (0.5

µL)

Procedure:In brief:

I. Add the adding components to the sterile, labeled 0.5 mL microcentrifuge tube on ice:

(d)autoclaved distilled H2O (78.5 µL)(e)PCR buffer w/ MgCl2 (16 µL)(f) dNTP 10 mM mixture (2 µL)(g)2 primer mix – primer I: 9877, primer II:

9899 (2 µL – 1 µL for each primer)(h)control DNA (1 µL)(i) Taq DNA polymerase (0.5 µL)

2. Cap tube and mix the contents briefly (centrifuge)/change the tips of the micropipette

3. Add mineral oil (50 µL) , which will overlay on the surface – the mineral oil acts as an agent to prevent evaporation, since there is only 0.5 µL of Taq DNA polymerase

4. Again, cap tube and mix contents briefly (centrifuge). Then collect the contents at the bottom of the tube with micropipette. Be careful not to include any of the mineral oil.

5. The following PCR amplification was performed for the average 30 cycles:

(a)Denature = 95°C (1 min.)(b)Anneal = 50°C (1 min.)(c) Extension = 72°C (1 min.)

6. Incubate for additional time 72°C (10 mins.) & until used, store at a below freezing temp/ of at least 4°C

7. Analyze the amplification products by agarose gel electrophoresis and visualize

by ethidium bromide staining, using appropriate molecular weight standards.

Results:The expected PCR product (200-300 bps) can be verified by running an agarose electrophoresis gel, which unfortunately, due to a lack

of time of his semester, has to be conducted at a later date.

PCR steps:(a) Denature = double-stranded DNA

molecule is heated (95°C for 1 min.) in the attempt to break the hydrogen bond of the complementary bases so that single strands result

(b) Anneal = 2 complementary oligonucleotide primers are attached to the ends of the single-strands and amplified at 50°C for 1 min. (the 2 oligonucleotide primers are designated so that they anneal to the opposite strands of the template DNA being amplified - the 3’ ends of the oligonucleotide primers face each other)

(c) Extension = these oligonucleotide primers are extended by Taq DNA polymerase in the presence of dNTP, which results in copies (templates) of the new DNA strand