Download BI:4224

Leethaniel Brumfield, III
Genetics – Spring 2007
Genetics
Table of Contents
Laboratory Experiments
Date
Lab Title
Pages
1-19-07
1-4
2-2-07
1: Laws of Probability & Chance
(Chi-Square Test)
2: Mitosis
(Onion Root Tip Experiment)
3: Mendelian Genetics
9 - 11
2-9-07
4: DNA Extraction from Plant Cells
12 – 16
2-16-07
17 - 24
3-16-07
5: Isolation & Purification of RNA
(from Gram-Negative Bacterial Cells)
6: Separation of Unknown DNA Fragments Using
Agarose Gel Electrophoresis
7: Applications of Recombinant DNA (Genome Analysis)
& Genetics Online Lab (DNA Databases)
8: Genetics of Human Taste Response
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
47 - 49
1-26-07
2-23-07
3-9-07
5-8
25 - 28
29 - 33
34 - 42
(No Procedure)
4-13-07
12: Restriction Digestion of DNA
50 - 54
4-20-07
13: Transformation of E. coli
55 - 60
(No Procedure)
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)
H
T
Observed Expected
(O)
(E)
46
50
54
50
100
(Total)
O-E
(O - E)2
(O–E)2/E
-4
4
16
16
0.32
0.32
X2 = ∑(O-E)2/E
df
(N-1)
1
1
0.64
Probability = 0.50 > 0.30
Chart II:
Tossing 2 Coins (2 Events Occurring Simultaneously – 96 Times)
Result
(N)
HH
Observed Expected
(O)
(E)
21
24
O-E
(O - E)2
(O–E)2/E
X2 = ∑(O-E)2/E
-3
9
0.375
df
(N-1)
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)
HTH
Observed Expected
(O)
(E)
26
27
O-E
(O - E)2
(O–E)2/E
X2 = ∑(O-E)2/E
-1
1
0.037
df
(N-1)
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)
NN
Observed
(O)
272
Expected
(E)
249
O-E
(O - E)2
(O–E)2/E
23
529
2.12
df
(N-1)
1
nn
60
83
-23
529
6.37
1
332
(Total)
100%
Probability = 5.99 > 9.21
X2 = ∑(O-E)2/E
8.49
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)
A
B
AB
O
Observed Expected
(O)
(E)
326
(41%)
343.9
82
(9%)
75.5
27
(3%)
25.1
404
(47%)
394.3
839
100%
(Total)
O-E
(O - E)2
(O–E)2/E
-17.9
320.41
0.931
df
(N-1)
3
6.5
42.25
0.559
3
1.9
3.61
0.143
3
9.7
94.09
0.238
3
Probability = 0.70 > 0.50
X2 = ∑(O-E)2/E
1.87
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 kinds
Dihybrid 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 wall
Chitin = 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 bath
3. decant supernatant & add cell lysis solution
4.incubate (@ 65°C/ 1 min) to lyse cells
5. cool (@ room temp/4 mins)
II. Protein-DNA Precipitation
1. add protein-DNA precipitation solution to cell lysate
2. put tube in ice bath
3. centrifuge (10,000 RPM/6 mins)
the precipitated protein & DNA forms a light pellet at tube bottom
III. RNA Precipitation
1. pour the RNA supernatant into another tube/leave precipitate proteinDNA pellet
2. add 95% isopropyl alcohol to supernatant
DNA 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% ethanol
5. 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 T1
1. 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 S1
1. 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 I
1. 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 K
1. serine protease w/ broad cleavage specificity isolated from the fungus
Tritirachium album Limber
2. 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 sequence
NCBI = National Center for Biotechnology Information
Human Genome Resources = OMIM (Online Mendelian Inheritance in Man)
Cytochrome P450 = a protein
Entry = the life science search engine
Getting into databases = Pubmed (NCBI- Molecular Database)
BLAST = Base Local Alignment Search Tool
Blastp = 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 proteincoding 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 genome
c. F = Characterizing the transcriptome and proteome present in a cell at a specific
developmental stage or in a particular disease state
d. 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 genome
f. F, C = Determining the function of human genes by studying their homologs in nonhuman
organisms
g. F = Developing a comprehensive two-dimensional polysaccharide gel electrophoresis map of
all proteins in a cell
h. 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 experimentation
k. S = Identifying a large collection of simple tandem repeat or microsatellite sequences to use as
DNA markers within one organism
l. 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 method
o. S = Sequencing individual BAC or PAC clones aligned in a contig using a shotgun approach
p. 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 twodimensional 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
1.
2.
3.
4.
5.
6.
7.
8.
Background:
Taste Test Paper (PTC):
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
PTC: ability to taste PTC is inherited/ the taste dimorphism attaches to a
pair of genes
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
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.
some people have a higher taste sensitivity threshold for PTC than others
(Blakeslee)
classification into tasters & nontasters is just a reflection of the bimodal
distribution of taste threshold
sodium benzoate (similar to PTC) use is limited because some believe that
its use is detrimental to health
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 Modalities
1. 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 senses
b. Heritability of this variation
II. Bitter
1. 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 antimalarial 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 & Umami
1. Both indicate the presence of calorically rich & essential
nutrients
2. Very different perceptually, but similar phylogentically
3. Umami taste – performed with (MSG) monosodium glutamate
4. 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. Sour
1. A basic taste quality that ignites an innate rejection response
2. Acidic stimuli are the unique sources of sour taste
3. 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. Salty
1. 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
Sodium
Benzoate Pink
Thiourea Yellow
Bittersweet
(Sour, Salty)
Taster
Taster #1
Female
(1st)
Taster #2
Female
(2nd)
Taster #3
Male
*Represents positive taste response =
1st
Note:
Female = same food, different drinks (grandparent, parents/siblings)
2nd Female = same food, same drinks (parents & siblings)
Male = same food, different drinks (parent/siblings)
NonTaster
“How Taste Response Is Hard-wired Into the Brain”
More on: Neuroscience, Brain Injury, Perception,
Nutrition Research, Disorders and
Syndromes, Behavior
Date: January 25, 2006
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.
Source: Cell Press
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 machine
hemoglobin = 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 pellet
3. 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 bottom
8. keep supernatant & transfer to a new tube
9. add 600 µL methanol, 150 µL chloroform, & 450 µL of
distilled H2O
10. vortex (1 min)/ spin @ 6000 RPMs (5 mins)
11. discard upper layer, leaving interface & lower layer
12. 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:
Antibodies - defend the body from germs
B. Contractile Proteins - responsible for movement
Enzymes - speed up chemical reactions/Example: Hemoglobin = carries
oxygen through the blood
D. Storage Proteins - store amino acids
A.
C.
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 boiled
Answer: 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°C
Answer: 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 Electrophoresis
Agarose 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
Couldn't you just dye?
have a high negative charge. When
DNA is placed on a field with an
Ethidium bromide is an intercalating dye, which
electric current, these negatively
means it inserts itself between the bases that are
charged DNA molecules migrate
stacked in the center of the DNA helix. One
toward the positive end of the field,
ethidium bromide molecule binds to one base. As
which in this case is an agarose gel
each dye molecule binds to the bases the helix is
immersed in a buffer bath. The
unwound to accommodate the strain from the dye.
agarose gel is a cross-linked matrix
that is somewhat like a threeClosed circular DNA is constrained and cannot
dimensional mesh or screen. The
withstand as much twisting strain as can linear
DNA molecules are pulled to the
DNA, so circular DNA cannot bind as much dye
positive end by the current, but they
as can linear DNA.
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 LB
Devon’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.
2.
3.
4.
5.
6.
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)
Cap tube and mix the contents briefly
(centrifuge)/change the tips of the micropipette
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
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.
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.)
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)
(c)
Anneal = 2 complementary oligonucleotide
primers are attached to the ends of the singlestrands 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)
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