Biology 3338A Lecture Notes - Lecture 1: Ian Wilmut, Camera Dolly, Xenopus

26 views3 pages

coffeemoose574

9 Jul 2016

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Developmental biology is about how the body is developed. If you consider hemoglobin, a geneticist studies how the gene is inherited, a physiologist asks how does hemoglobin binds oxygen (at what temperature etc ) We know that animals start off as one cell which contains one set of dna and then it develops into all the different types of cells etc . Single cell organisms can change and have multiple cell cycle but not necessarily develop. We focus on multi-cellular organisms in this course. Usually we have gametes and end up with a fertilized zygote which divides. As it divides you end up with different structures (ie gastrula etc ) Different germ layers give rise to hundreds of different cell types, do all these cells have the same dna? yes. Well if they have the same dn a how do you get those cells in the right place and the right time.

Related Questions

In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that

	DNA is a double helix.
	DNA may be taken up by bacterial cells and be active.
	DNA is present in bacteriophage T2.
	eukaryotic cells may be genetically transformed.
	DNA binds to the cell wall of the bacterium.

In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.

	temperatures high enough to kill cells
	enzymes that hydrolyze proteins
	enzymes that hydrolyze DNA
	enzymes that hydrolyze polysaccharides
	enzymes that hydrolyze RNA

Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?

	Pure RNA from S cells
	Cell-membrane extracts from S cells
	A mixture of pure RNA and protein from S cells
	Cell-free extracts from S cells
	Pure protein from S cells

A-T base pairs in a DNA double helix

	form three hydrogen bonds with each other.
	are more heat-stable than G-C base pairs.
	are of a substantially different length than G-C base pairs.
	are not accessible to DNA binding proteins.
	are chemically distinct from G-C base pairs.

If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?

	27
	50
	0
	73
	23

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only ¹⁵N and the second only ¹⁴N, this result would have indicated that replication was

	unidirectional.
	dispersive.
	conservative.
	bidirectional.
	semiconservative.

10.

The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3â² hydroxyl group (âOH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.

	The replication complex will no longer bind to origins of replication.
	The DNA polymerase will only incorporate acyclovir molecules.
	Acyclovir will bind single-strand binding proteins.
	DNA polymerase will not be able to link a successive nucleotide.
	Acyclovir will block the activity of the DNA helicase.

11.

Which of the following does not demonstrate the stability of the DNA double helix?

	Single-strand binding proteins are required to keep the strands apart.
	Paired bases form hydrogen bonds.
	High temperatures are required to denature it.
	DNA synthesis requires energy.
	DNA helicases require ATP.

12.

What effect would a primase inhibitor have on DNA replication?

	DNA polymerase would not be able to add bases to the DNA.
	Primase would be blocked from joining the DNA replication complex.
	Primase would not be able to provide primers for DNA polymerases.
	DNA polymerase would not be able to use DNA as a template.
	There would be no effect on DNA replication.

13.

To replicate their DNA in a timely manner, most eukaryotic chromosomes

	normally have uncoiled DNA.
	have multiple origins of replication.
	contain linear molecules of single-stranded DNA.
	have two replication forks that move in the same direction.
	are composed entirely of DNA.

14.

Which statement about DNA replication is false?

	Okazaki fragments are synthesized as parts of the leading strand.
	Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
	The sliding clamp increases the rate of DNA synthesis.
	Replication forks represent areas of active DNA synthesis on the chromosomes.
	Ligases and polymerases function in the vicinity of replication forks.

15.

In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.

	leading; telomerase
	lagging; telomerase
	lagging; DNA polymerase I
	leading; DNA polymerase I
	leading; replicase

16.

A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?

	DNA polymerase is constantly active.
	Primase is able to extend the telomeres.
	The mismatch repair system is extra efficient.
	The telomerase gene is being expressed.
	The cells do not need the ends of their chromosomes.

17.

If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?

	The replication complex excises the incorrect nucleotide.
	The excision repair proteins remove the incorrect nucleotide.
	The mismatch repair system scans for base-pairing mismatches.
	There is no correction, because nucleotide errors by DNA polymerases are not repaired.
	DNA ligase connects the correct nucleotide.

18.

Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.

	1, 2, 3, 4
	2, 1, 3, 4
	3, 4, 2, 1
	2, 4, 1, 3
	4, 2, 3, 1

19.

Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.

	64
	32
	12
	128
	6

20.

When double-stranded DNA is heated to temperatures above 90Â°C, it denatures. Denaturation is a process that

	breaks covalent bonds.
	hydrolyzes DNA.
	separates double-stranded DNA into two single strands.
	causes new hydrogen bonds to form.
	irreversibly destroys the double helical structure.

Question 9

Codominance is a form of inheritance in which two different allelesfor a gene are both expressed, and neither allele is dominant overthe other. One example of codominance is fur color in cattle; thealleles for red fur and white fur are codominant. If a homozygousred cow and a homozygous white bull mate, what would you expect tofind in their offspring?.912.l.16.2>

They will all have white fur.

Their fur will be a mixture of red and white hairs.

Their fur will be a lighter shade of red.

They will all have red fur.

Question 10

If a tRNA molecule has an anticodon which reads AUG, what will itmatch up with and what amino acid is it carrying?
.912.l.16.5>

	U		C		A		G
	UUU	Phenylalanine	UCU	Serine	UAU	Tyrosine	UGU	Cysteine	U
U	UUC	Phenylalanine	UCC	Serine	UAC	Tyrosine	UGC	Cysteine	C
	UUA	Leucine	UCA	Serine	UAA	Stop	UGA	Stop	A
	UUG	Leucine	UCG	Serine	UAG	Stop	UGG	Tryptophan	G
	CUU	Leucine	CCU	Proline	CAU	Hisitidine	CGU	Arginine	U
C	CUC	Leucine	CCC	Proline	CAC	Hisitidine	CGC	Arginine	C
	CUA	Leucine	CCA	Proline	CAA	Glutamine	CGA	Arginine	A
	CUG	Leucine	CCG	Proline	CAG	Glutamine	CGG	Arginine	G
	AUU	Isoleucine	ACU	Threonine	AAU	Asparagine	AGU	Serine	U
A	AUC	Isoleucine	ACC	Threonine	AAC	Asparagine	AGC	Serine	C
	AUA	Isoleucine	ACA	Threonine	AAA	Lysine	AGA	Arginine	A
	AUG	Methionine/Start	ACG	Threonine	AAG	Lysine	AGG	Arginine	G
	GUU	Valine	GCU	Alanine	GAU	Aspartate	GGU	Glycine	U
G	GUC	Valine	GCC	Alanine	GAC	Aspartate	GGC	Glycine	C
	GUA	Valine	GCA	Alanine	GAA	Glutamate	GGA	Glycine	A
	GUG	Valine	GCG	Alanine	GAG	Glutamate	GGG	Glycine	G

It is carrying arginine and will match with a CGU codon on themRNA.

It is carrying methionine and will match with a TAC codon on theDNA.

It is carrying methionine and will match with a UAC codon on themRNA.

It is carrying tyrosine and will match with a UAC codon on themRNA.

Question 11

While the DNA in a human skin cell was being replicated, a singlebase was miscopied. What will be the most likely result of this forthe cell in which it happened?.912.l.16.3>

All the proteins the cell creates from the miscopied strand willdo different jobs than the old ones.

If the new sequence codes for the same amino acid as theoriginal cell, it will function normally.

Both new DNA strands will end up together in a new cell, and theinaccurate one will be discarded.

Any miscopied DNA will be replaced with an accurate DNA copyonce the cell divides.

Question 12

A ferret's haploid number of chromosomes is 20. How would thenumber of chromosomes in the ferret's body cells compare to thenumber of chromosomes in its gametes?.912.l.16.17>

Its body cells would have 20 chromosomes, and its gametes wouldhave 40 chromosomes.

Its body cells and gametes would both have 40 chromosomes.

Its body cells and gametes would both have 20 chromosomes.

Its body cells would have 40 chromosomes, and its gametes wouldhave 20 chromosomes.

Question 13

A body cell is in the longest stage of its life cycle. The cellgrows, synthesizing proteins and increasing in size. Eventually,the cell will grow too large to carry out normal activities. Whichof the following is the best conclusion you can make about the lifecycle of this cell?.912.l.16.14>

The cell is close to the end of its life cycle, and a chemicalsignal will initiate cell death.

The cell is ready to undergo mitosis, and a chemical signal willsend the cell to prophase.

The cell is in the G1 phase of interphase, and a chemical signalwill send the cell to the S phase.

The cell is in the S phase of interphase, and a chemical signalwill move the cell to the G2 phase.

Question 14

While mRNA strands are being created a sequence is sometimesmiscopied. What is the best possible outcome for the cell shouldthis take place?.912.l.16.5>

The miscopied sequence codes for the same amino acids as theoriginal sequence.

The new sequence creates a protein that serves a differentfunction from the original.

The ribosomes will correct the mistake before the tRNA matchesan amino acid to it.

The mRNA will only be used to create non-critical proteins forthe cell.

Question 15

Over the last several decades, the scientific community hasgathered a large amount of information regarding genetics andgenetic variation. What are two main sources that lead toincreased genetic variation?.912.l.15.15>

Selective breeding

Gamete mutations

Recombination

Genetic drift

salmonlouse680

2.Conclusion:

What have you learned from your study?

Interpret your data/ results:

Do your results support of disprove your hypothesis?

Compare and contrast your result with those of previous research.

Significance of the results: âBig Pictureâ

What is the global impact of your results?

What questions remain unanswered?

What kinds of experiments can come out of your study?

i have lab for biology and i need help with the Lab Report

the lab name is Bacterial Transformation Lab

1. Introduction:

Introduce the background and the nature of the problem your experiment is investigating.

Summarize the current knowledge of your subject.

Use examples of other research that have examined the same or a similar topic as your experiments.

Ending your introduction:

State your hypothesis

State your predictions

Give a brief statement about how you tested your hypothesis.

please help me with it is important for my lab class tomorrow

that is the lab i put the info please help me for the intro and for number 2 please be clear please

i would be happy if you help me it is due tommorrow i realy need your help on the intro and number 2 please be clear and show clear work hope you do it because it is due tomorrow

Bacterial Transformation Lab

Genetic engineering is one of the newest applications of basic knowledge in biology. It involves the isolation of specific genes from one organism and the insertion of those genes into a second organism of the same, or even different, species. Remember that a gene is a piece of DNA which codes for a product, usually a protein, which gives an organism a particular trait. The development of these techniques has generated considerable excitement and answered many questions, but it has also raised questions. Many scientists and others see genetic engineering as offering opportunities to make life better, for example, by exploring how genes are regulated, by curing diseases of genetic origin, by adding desirable traits to agricultural crops such as the ability to fix nitrogen in corn and pest resistance in cotton, and by genetically altering bacteria with genes enabling them to digest oil spills.

For almost 100 years, it has been possible to isolate DNA from cells, but only in the last 40 years or so has its function been understood. Your textbook discusses the work of such investigators as Fred Griffith and Oswald Avery, who demonstrated that if DNA was isolated from a pneumonia-causing strain of bacteria and added to cultures of nonvirulent bacteria, the recipient cells were transformed into virulent types that caused pneumonia. These transformation experiments were the first to show that genes could be artificially transferred in the laboratory. Since the time of those experiments (Avery's work was done in the 1940s), our knowledge of gene structure, function and control has increased astronomically. The application of this knowledge has created the field of genetic engineering.

Modern genetic engineering techniques depend on three critical factors:

A suitable host organism in which to insert the foreign gene;

A vector to carry foreign genes into the host; and

A means of isolating the host cells that have taken up the foreign gene.

The host organism

The most commonly used host organism is the gut bacterium Escherichia coli (E. coli). Its chromosome is a single large circular DNA molecule, called the genomic DNA, containing about 4 x 10⁶ base pairs. Not every cell will take up foreign DNA when exposed to it, but E. coli grows rapidly, dividing in less than 20 minutes, so that a single cell can give rise to a billion genetically identical descendants in a matter of 10 hours. Thus, a few transformed cells can provide a large number of offspring within a day. The strain we are using is K-12.

The vector

A vector is a DNA molecule used as a vehicle to carry genes from one organism to another. This DNA molecule can be modified by gene-splicing techniques using restriction enzymes and ligases so that it carries the genes of interest (essentially a cut-and-paste operation). Sometimes viruses are used as vectors, but for E. coli, plasmids are often used. A plasmid is a small circular DNA molecule containing 1000 to 200,000 base pairs. Plasmids exist naturally in many strains of E. coli and other bacteria. Plasmids replicate as the cell grows, independent of replication of the genomic DNA. These plasmids are transmitted to each of the product cells during cell division. Plasmids carrying genetic information that is beneficial to the host cell are maintained in a given population by natural selection. Plasmids frequently carry genes that confer antibiotic resistance, an obvious benefit.

When E. coli cells are exposed to plasmids or any other foreign DNA, some of the cells, called competent cells, will take up the DNA. We can induce cells to become competent by treating them with divalent cations, such as Ca⁺⁺, followed by a brief exposure to high temperature, called a heat shock treatment. Exactly how this treatment causes cells to become competent is not understood, but it is described as making the membranes "leaky" so that large DNA molecules (like plasmids), which would not normally pass through the cell membrane, can enter the cell. This treatment does not make all cells competent, but it greatly increases the number of competent cells. Once in a cell, the plasmids replicate and are passed to the product cells during cell division.

The selective medium

The last requirement for genetic engineering is a means of isolating host cells that have taken up the gene of interest. The techniques used depend on the growth requirements of the host organism. For example, each species of bacteria has particular nutrient requirements. The composition of the growth medium (agar or broth) can be modified to allow growth of only certain bacteria -- for example, by adding or removing nutrients or other chemicals from the medium mixture. This modified medium is called a selective medium, because it selects for the growth of only some species. By introducing an antibiotic such as ampicillin into the medium in which the bacteria are grown, you will select for growth of only those bacteria carrying the antibiotic-resistance gene on newly acquired plasmids. In a matter of days, billions of antibiotic-resistant cells, all with the plasmid, can be grown.

Genetic engineers use a modification of this technique to insert a foreign gene into E. coli. In simplified terms, the plasmid (already carrying an ampicillin resistance gene) is modified by inserting the gene of interest. The gene of interest has been isolated from another source; for example, the gene for human insulin has been isolated from the human genome. The modified plasmid, carrying the inserted gene, is then mixed with competent E. coli and the bacteria are placed in a medium containing ampicillin. Only those cells that have the plasmid will grow in the medium, and they are also the cells carrying the insulin gene (in their plasmids). These selected cells can now be used to manufacture insulin, which can be used for insulin therapy by diabetics.

The genes

In the following experiment, you will insert a plasmid into E. coli. The plasmid you will use carries the gene bla, which codes for a protein (beta-lactamase) that is secreted by the bacteria, inactivating the ampicillin in the agar and allowing for bacterial growth. The plasmid also carries an indicator gene. The indicator gene we are using was isolated from a bioluminescent jellyfish, Aequorea victoria. It codes for Green Fluorescent Protein (GFP).

The GFP gene has been modified to include a regulator system that can be used to control the expression of the gene. Gene expression in all organisms is carefully regulated to allow for adaptation to differing conditions and to prevent wasteful overproduction of unneeded proteins. The genes involved in the breakdown of different food sources are good examples of highly regulated genes. For example, the sugar arabinose is both an energy source and a carbon source for bacteria. The bacterial genes that make digestive enzymes to break down arabinose for food are not expressed when arabinose is not in the environment. But, when arabinose is present, these genes are turned on. When the arabinose runs out, the genes are turned off.

Arabinose initiates transcription of these genes by promoting the binding of RNA polymerase. In the genetically engineered pGLO DNA, some of the genes involved in the breakdown of arabinose have been replaced by the jellyfish gene GFP. When bacteria with this plasmid are grown in the presence of arabinose, the GFP gene is turned on and the bacteria glow green in UV light. When arabinose is absent form the growth medium, the GFP gene remains turned off and the colonies appear white. This is an example of the central molecular framework of biology in action: DNA ? RNA ? protein ? trait.

In summary, you will render E. coli cells competent to take up plasmid DNA. The plasmid carries two identifiable phenotypic markers:

The beta-lactamase or bla gene codes for resistance to the antibiotic ampicillin. We will use this to select for bacteria that have taken up the plasmid.

The pGLO gene codes for Green Fluorescent Protein (GFP), which is switched âONâ if arabinose is present.

Safety Procedure

For your own safety and the safety of others, it is essential that you follow all instructions about handling bacteria and disposal of materials. When you finish work for the day, swab your table with disinfectant. WASH YOUR HANDS thoroughly with soap before leaving the lab for any reason, and at the end of the day after you have cleaned your table. Report any spills to your instructor.

List of Materials

Prepared agar plates (2 LB, 2 LB/amp, 1 LB/amp/ara)

Transformation solution (50 mM CaCl₂, pH 7.4)

LB broth

1 pack of inoculation loops

Sterile transfer pipets

1 foam microtube holder w/ tubes

E. coli starter plate

1 red biohazard bag for loop disposal (common use â Loops/pipets only - NO PAPER!!!)

beigefox320

Biology 3338A Lecture Notes - Lecture 1: Ian Wilmut, Camera Dolly, Xenopus

Document Summary

Get access

Related Documents

Biology 3338A Study Guide - Midterm Guide: Desmosome, Zygosity, Mothers Against Decapentaplegic Homolog 4

Biology 3338A Final: 151-Page Final Exam Notes (Sections 1-14) - with Pictures and Explanations!

Biology 3338A Lecture Notes - Lecture 1: Roslin Institute, Cell Nucleus, Udder

Related Questions