[Final] Lecture 36 12/8; Review

Review sheet on blackboard! Contains the main concepts that will be on the test.

(6:00) Complete linkage.

(17:15) Three factor cross mapping

• (30:10) Mapping distance

(34:40) Chi squared analysis

(38:30) Meiosis

(39:05) Lyon Hypothesis


This is the review he posted on blackboard.

BIOL 3210
Final Exam Review Sheet

This review sheet will give you potential topics for problems/ short answer/ essay questions. All of the material covered during the semester is subject to multiple choice questions.

1.) How to perform monohybrid and dihybrid crosses
2.) How to perform trihybrid crosses using the forked-line or branch diagram
3.) How to perform chi-square analysis and determine whether or not to reject or fail to reject a hypothesis
4.) The application of the binomial theorem in order to calculate the probability of an event happening
5.) The various unusual patterns of inheritance and how those patterns affect the passage of genes from one generation to the next
6.) How to perform linkage mapping, calculate mapping units between genes, calculate interference and coefficient of confidence
7.) Being able to draw all four structures of the bases and the structures of deoxyribose and ribose
8.) Understanding the experiments of Griffith. How were they performed and what were the results
9.) Understanding the experiments of Avery, MacLeod and McCarty. How would the results have been different if RNA was the genetic material and not DNA.
10.) Understanding the Hershey-Chase experiments. How were they performed and what were the results. How would the results have been different if RNA was the genetic material and not DNA?
11.) Describe the experiments that showed that Hfr strains could be used to determine the relative order of genes on the prokaryotic chromosomes
12.) Understand how the fluctuation test was performed. What were the two hypothesis that were being tested? What were the results that were obtained and how would the results have been different if the other hypothesis was correct?
13.) Use of pedigree analysis to follow a trait through generations within a family. Using this technique to understand the mode of inheritance as well as to determine the genotype or potential genotypes for individuals in a family
14.) Describe the Linderberg –Zinder experiment and the nature of transduction
15.) Understand the evidence that Watson and Crick used in order to describe the double helical nature of DNA
16.) Be able to describe the experiment that first showed that genetic recombination required physical interaction between cells
17.) Be able to describe the complete life cycle of a bacteriaphage
18.) Understand how gene duplication functions to allow for the formation of new genes
19.) Know the sequence elements found in the eukaryotic promoter and how their importance was established
20.) Be able to describe the nature of the regulation of gene expression for the lac operon in the presence of: (a) lactose; (b) no sugar; (c) lactose and glucose
21.) Be able to describe the nature of the gene regulation for the GAL genes in the presence of: (a) galactose; (b) no sugar; (c) galacotse and glucose
22.) Be able to describe the nature of the regulation of the trp operon in the presence and absence of tryptophan
23.) Understand how compounds are tested to determine if they are in fact carcinogens
24.) Know the levels at which eukaryotic gene regulation can take place: be able to give an example for each level
25.) Be able to describe the first two levels of chromosome packing and the key pieces of evidence that supported these two models
26.) Be able to describe tumor suppressor genes and oncogenes. Be able to give an example of each
27.) Be able to describe the process of nucleotide excision repair
28.) Describe the lyon hypothesis and the data that supports the lyon hypothesis
29.) Be able to describe the differences between maternal effect and organelle heredity
30.) Be able to describe how a monosomy or trisomy arises and give an example of a disease caused by each

Lecture 35, 12/3; Cancer

We have been talking about genes effect on cell cycle regulation.  We are now going to talk about how they directly impact the cell cycle.  

Two types of genes that fall into each category

1.) Proto-oncogenes.  Normally function to promote the growth of the cell and cell division.  These genes are not always on.  When a proto-oncogene is mutated it typically is a constituitive mutation.  The gene is constantly telling the cell to grow and divide.
A mutated proto-oncogene becomes an oncogene.

• EX. ras family genes.
• Normally proteins from these genes are GTP binding proteins.  GTP binding protein bound to GTP is on.  Over time, GTP is converted to GDP and the protein becomes inactive.  Eventually GDP is released and GTP is rebound (It's a cycle).  When the gene is mutated it loses the ability to convert GTP to GDP.  Because of this, the GTP protein becomes permenantly bound to GTP and is always on. 
  
• SO . . . the ras family genes normal role is promoting cell division.  When the mutation is made and the gene is on all the time (cell division occurs at high rate).  Often, what causes this is a point mutation.  Mutations in ras family genes found in 40% of cancers. 
  

(10:00) 2.) Tumor supressor genes.  Normal function of a tumor supressor gene is to slow/stop cell growth.  Often, tumor supressor's help trigger apoptosis. Mutated tumor suppressor genes = loss of the ability to stop cell growth. 

• EX. P53 gene.  A mutation in the p53 gene is found in 50% of cancers.  
• P53 is constantly transcribed, but the protein is rapidly degraded which results in low levels of the protein.  Normal roll of p53 is that it is a transcription factor and it regulates the expression of 50 genes.  It is both a positive and negative regulator. 
  
• The expression of p53 is turned up in certain situations such as repairing DNA damage caused by UV light.  The normal roll is to arrest the cell cycle to allow time for DNA repair.  If we can't fix the damage we trigger apoptosis.
• SO . . . when p53 is mutated the cell cycle con't arrest and the cell begins to rapidly accumulate mutations.  In addition you can't trigger apoptosis as easily. 
  

(20:35) Predisposition to cancer can be inherited.  1 to 2 % of cancers are hereditary.  Overall there are 50 forms of hereditary cancer.

• Typically the genes responsible for these cancers are autosomal dominant.  By themselves they are not sufficient to cause cancer, however. 
  
• One mutated allele = no cancer.  When you mutate the second allele in a heterozygote head down the pathway to cancer (loss of heterozygosity. 
  
• After loss of heterozygosity the cell will form a number of additional mutations then you see tumor formation. 
  
• (25:40) EX.  Familial adenomalus polyposis.
• individuals with this mutant syndrome have an extra copy of the APC gene which is a tumor supressor. 
  
• THis mutation will cause a number of cells in the colon to lose cell cycle control.  The result of this is the formation of 100 to 1000 rectal polyps.
• Polyp can transition to a malignant tumor.  This happens as you accumulate a second mutation in a ras family gene.  Finally you have a mutation in the DCC gene.  The end result is a late stage adenoma.

(31:25) Viruses can contribute to cancer.  This started with the understanding that viruses can cause cancer in animals.

• in animals it is caused by a retrovirus which is broadly labeled as an acute transforming virus. 
  
• Discovered in 1910 by Francis Rous as he was looking at sarcoma's in chickens (tumores in muscle bone or fat).  
• What he did: Injected chickens that did not have a tumor with "ground up tumor" and found that they developed this tumor.  Referred to as the Rous virus.

(36:45) How does this work?

• a virus enters a cell and their RNA is converted to DNA by reverse transcriptase.  the DNA integrates itself into the genome of the host cell. 
  
• Two ways this can cause cancer.
• 1.) if the DNA is inserted near a proto-oncogene it can cause the proto-oncogene to become an oncogene.
• 2.) The viral DNA can ring with it a proto-oncogene which becomes an oncogene over time. 
  

15% of all human cancers are caused by viruses.  2nd leading cause of cancer. 

(40:00) Common human viruses associated with cancer: Human papilloma virus (HPV), epstein Bar virus, human t-cell leukemia virus.  The disease itself is not sufficient to cause the cancer, you need additional mutations to happen over time.

A closer look at HPV - leading cause of anogenital cancer in the world. 

• Can lead to Cervical cancer.  90% of all cervical tumors have HPV DNA

• The virus carries two proto-oncogenes (genes that help it cause cancer)
  

• E6 - binds p 53 and inactivates it
  
• E7 - functions to stimulate cell growth and division.
  

(46:15) Environmental Agents that cause cancer.
Any substance that damages DNA can be a carcinogen.
Chemicals and radiation are the primary environmental contributers to cancer.  The most common environmental carcinogen is tobacco smoke.  Compounds within tobacco smoke interact with both the p53 gene and the ras genes.
-->30% of all cancers are caused by tabacco smoke. 

Diet can also impact cancer.  Colon, prostate and breast cancer is higher in individuals that eat high fat and high red meat diets.  It is suggested that hormones in the meat or byproducts in them are a contributing factor. 

UV light is another major contributor to environmental causes of cancer.  It causes DNA lesion which leads to cancer.

Lecture 34, 12/1; Chapter 18, Cell Cycle Regulation and Cancer

Chap 18
Cell Cycle Regulation and cancer.

-Cancer is the 2nd leading cause of death in western world.  1 in 3 people will be diagnosed with cancer at one point in their life.
-1,000,000 new cases of cancer in the U.S. each year.
-500,000 deaths attributed to cancer

Cancer is a genetic disease. 
-The genomic alteration associated with cancer range from single nucleotide changes to large deletions.

• Vast majority of these mutations arise in somatic cells.
• only 1 % of cancers are linked to germline mutations which predispose an individual to cancer.

Most cancers are a result of multiple mutations (6-10 avg) as opposed to a single mutation.

(9:20) Common traits of cancer.

• cell proliferation - abnormal unchecked growth
• Metastisis - cells have the ability to spread to other parts of the body

typically two broad categories of tumors

• benign - tumors that are localized to one region and are not metastisis.
• malignent - tumor which has the ability to spread to other areas of the body

all tumors arise from a single cell.

• begins to accumulate mutations and eventually starts to grow uncontrolably.

(15:25) The idea that "all tumors arise from a single cell" is supported by x inactivation data. 

• cells in a single tumor all share a common inactivated X chromosome.
• Because X inactivation is random the likelihood that all cells in a tumor share an inactivated X chromosome by chance is extremely low.

(18:05) Broadly: Cancer is a multi-step process.

• This idea that "cancer is a multi-step process" is supported by the following data.
• 1.) There is a definite increase in the incidence of cancer as age increases.
• 2.) Also when looking at victims of the atomic bombing in WWII we see that 5-8 years after the bombing there was an increase in the rate of cancer for survivors of the initial blast.

(21:30) Example of the multi-step process: Cervical Cancer.

• In the normal cervix a number of cells, after time, become quiescent cells (a cell that has entered Go phase and stopped dividing). 
  
• However, a number of basal cells exist in the cervix (not quiescent) which are actively dividing or have that ability.
• Occasionally basal cells form mutations.  When these mutations happen they can form a dysplasia (tumor area).  This is early cervical cancer and easily treatable.  Untreated the cells in the dysplasia will develop additional mutations over time.  With time the dysplasia becomes carcinoma which is much more difficult to treat.

(32:15) Genetic causes of cancer
Some cancer cells contain genetic defects affecting DNA repair and genomic stability. 
-On average, cancer cells have higher rates of mutations than normal cells. This suggests problems with DNA repair.
-Mutator phenotype - a cell more prone to mutation.

(34:45) Two types of cancer caused by mutations in the DNA repair system (inheritable forms of cancer)
-Xerodermo pigmentosum - very susceptable to the effects of UV light.  These indivuduals have a mutation in one of the 7 genes involved in nucleotide excision repair. 
-Hereditary nonpolyposis Colon cancer - this is an autosomal dominant allele (will run through families very strongly).  Mutation is involved in genes involved in mismatch repair system.

(39:05) Cancer is often caused by mutations that affect the cell cycle.
One of the hallmarks of cancer is that cells grow and divide at a greatly increased rate.
The cell cycle control system helps to prevent the uncontrolled growth of cells.

• During G1 the cell makes a decision about whether or not to continue to divide. A cell that stops actively dividing enters the Go phase.  In the Go phase cells are metobolically active . . . but not dividing.  A cell in the Go phase is referred to as a quiescent cell.
• Another one of the hallmarks of a cancer cell is that it skips Go phase.  If the cell does enter Go phase it passes through very quickly.
• (45:30) Normal cells will exit the Go phase in response to environmental signals --> A signal transduction pathway accomplishes this. 
  In a cancer cell, the genes in that signal transduction pathway are mutated so that the signal pathway is always on (tells it to grow more). 
  
• The cell cycle has checkpoints to help prevent this from happening: G1s checkpoint, G2 M checkpoint and the M checkpoint.
• BUT mutations in the genes for these checkpoints have an association with cancer.
  

(50:25)Apoptosis - programmed cell death.  A pathway  a cell can activate when its too damaged to continue.  When it is activated the cell self-destructs.

• BUT mutations in the genes for apoptosis are also associated with cancer.
  

Lecture 33, 11/24; Chapter 17 Regulation of Gene expression

Chapter 17 Regulation of Gene expression in Eukaryotes

Positive induction and catobolite regulation in yeast.
What are we looking at: The regulation of genes involved in galactose breakdown in yeast.

• In the absence of galactose genes are off.

• Presence of galactose the genes are expressed. (energy is conserved)
• Galactose genes are only expressed when glucose levels are low - catoblite repression.
• A mutation in the GAL 4 gene prevents the activation of the galactose genes even under appropriate conditions.
  

(6:00) What does GAL 4 do?

• GAL 4 functions as an activator.
• When galactose is present GAL 4 turns on the galactose genes.  So . . . when you mutate it and it can't function you will never see activation.
  

Catabolite repression - requires an enzyme.  This is opposed to an E. coli in the lac operon where catabolyte repression is based on cyclic amp levels.

(8:30) We're going to look at the regulation of GAL 1 to GAL 10

• upstream of both genes - you have a DNA sequence called UASg (upstream activator sequence).  It is a 170 BP sequence that acts as an enhancer.
• within the UAS there are 4 binding sites for the GAL 4 protein (the activator protein).  These sites are always bound by GAL 4 protein.  The activator is always present HOWEVER, the genes are not always expressed.
• (12:00) How is it that they are not expressed?
• A second protein called GAL 80 protein can physically interact with the GAL 4 protein. (form a protein protein complex)
  
• GAL 80 protein blocks the portion of the GAL 4 protein which activates transcription.
  
• GAL 80 protein is always bound to GAL 4 protein
• How this works: To induce . . . galactose, bound to a phosphate group, binds to the GAL 4/GAL 80 complex.  In doing so alters the interaction between GAL 4 and GAL 80 in such a way as to expose the region of GAL 4 that activates transcription. 
  
• In the presence of glucose a protein kinase (an enzyme which phosphorilates things) goes to the promoter to repress transcription.

(18:40) Post transcriptional regulation of gene expression.  (side note review: Eukaryotic gene regulation happens at different points)

• often these processes occur during mRNA modification (modification being: splicing, 5' capping and polyadenylation).  You have this period of time (post transcription, pre translation) where you alter structure of mRNA which opens a window to allow for altering expression of a gene.  ONe of the major forms of this is alternative splicing.
  
• Alternative splicing: splicing is the process of removing introns from mRNA molecules.  If you change the splicing pattern (don't remove an intron) you will change the protein that an mRNA can make.
• What this does: Alternative splicing allows the proteone (the number of proteins in a cell) to be greater than the genome (number of genes in the cell).  It is proposed that this is a common phenomenon in vertebrates.  IN fact 30 to 60% of all human genes are alternatively spliced. 
  
• While the genome has 25-30,000 genes, the proteon has 100,000's of proteins. 
  

(25:10) How alternative splicing impacts cell function.
Cochlear hair cells and hearing. 

• Broadly: Hair cells beat to allow the air to hear different frequencies.  This is controlled by the alternative splicing of the mRNA.
  
• SLO gene contains 8 different alternative splicing sites.  This gene can be spliced into 500 different mRNA's.
• Each different protein is responsible for a different frequency.  Essentially you have used alternative splicing to create proteins to function for a broad cellular function.
  

(30:40) Use of a regulatory RNA to control gene expression.

• There is a short, 21 nucleotide, RNA that represses the translation of certain RNA molecules.  Similar RNA's have been found in the nucleus - which specifically bind and repress genes at the level of DNA. Ex. - RNAi and PTGS
  
• RNAi (RNA interference) in animal cells. Post transcriptional gene silencing (PTGS) in plants. 
  
• The process begins with the creation of a double stranded RNA 70 BP long.  It is then processed by a protein called dicer protein.  The dicer protein cleaves the RNA to 21 nucleotides (the result is referred to as an siRNA or short interfering RNA)  siRNA Binds to a protein called "RNA Induced Silencing Complex".  RNA binds to complimentary mRNA molecules and turns off their translation.  
  
• (36:20) miRNA or micro RNA - 19-24 nucleotides. Binds to a 3' untranslated region of an mRNA to turn it off.

Turning off the translation is valuable as it is critical to development and helps the cell ward off viruses.  RNA interference (discovered within past 5-10 years) is important and is being thought of as a potential theraputic agent.  Introduce RNA complimentary to a mutated gene and it will seek out hte bad RNA's to turn them off.

Lecture 32, 11/21; Steps to Transcription in Eukaryotes

Transcription in Eukaryotes requires multiple steps

• It begins with chromatin remodeling. Within the structure of the chromosome we have regions of both euchromatin and heterochromatin (these regions are highly condensed and genes in these regions are not expressed). Chromatin remodeling primarily occurs to regions of euchromatin.
• Broadly: Chromatin remodeling uses a series of proteins to weaken the interaction between the histones and the DNA.
• All of these proteins have ATPase activity - they all convert ATP --> ADP (requires energy).
  
• (4:20) One such protein is the SWI/SWF complex.
• a complex of 11 subunit proteins. One of the subunits binds DNA. Another of the subunits is the ATPase subunit.

(5:35) How chromatin remodeling occurs

• Modify the interaction between DNA and Histones. This causes the Histone to slide down the DNA. (REMEMBER: We have DNA wrapped around the histone protein, that DNA that is interacting with the histone protein will be very inaccessbile.) As the histone moves the DNA that was attached to the histone, becomes exposed.
  
• The protein complexes (like the SWI/SWF complex) function to physically pull the DNA away from the histone, thus allowing access.
  

(9:10) Another way to accomplish chromatin remodeling.

• Catalyzed by histone acetyl transferase (HAT). These enzymes can add acetyl groups to the histones. Weaken the ineraction between the histone and DNA.
• When transcription is finished histone deacetylase (HDAC). They remove the acetyl groups and restore normal chromatin structure.
• Typically remodeling process starts just before the promoter for a gene and ends at the end of a gene. The portion that is going to be copied is the only thing that is opened up - to do this an insulator element is used.
  
• Insulator element binds to proteins to prevent the spread of remodeling.

(15:05) After remodeling the process moves on to the Assembly of the basal transcription complex

• Eukaryotes have multiple RNA Polymerases that transcribe different things
• RNA POL I is responsible for the transcription for rRNA.
  
• RNA POL II copies mRNA (mRNA is a major portion of the cell) and snRNA (snRNA does splicing).
  
• RNA POL III transcribes tRNA and 5SrRNA.
• each polymerase recognizes different promoter sequences.
  

(20:05) How RNA POL II initiates transcription.

• A number of proteins will come together to form a pre-initiation complex on which RNA POL II lands. The pre-initiation complex is recognized by the RNA POL II as a place to land and "act". What is involved in making the pre-initiation complex.
  
• TFIID (TF=transcription factor, II=RNA POL II, D= just a differentiating letter)
• Recognizes and binds the TATA box. A number of other proteins bind TFIID (there are many but they are not important).
  
• Ultimately RNA POL II binds this complex. After binding, RNA POL II leaves the TATA box and starts transcription at the basal level. Enhancers or silencers can alter the level of transcription.
  
• Activator proteins can enhance transcription 100 fold. (Whatever the basal level is, multiply by 100). The activator protein binds to enhancer DNA sequences to accomplish this.
  

(26:00) What does an activator protein look like? (It has two regions or domains)

• DNA binding domain - binds to DNA. Specifically an enhancer sequence.
• trans-activating domain: 30 -100 amino acids. It interacts with other transcription factors or RNA POL. If functions to increase the level of binding by RNA POL. In order to stimulate transcription you need to have RNA POL bind more frequently.

(30:00) SUMMARY
To start the process we do chromatin remodeling and open up the DNA so that it is accessible to RNA POL. After we accomplish that we assemble our complex then bind RNA POL, and in order to enhance transcription we utilize enhancer proteins.

Lecture 31, 11/19; Chapter 17, Regulation of Gene Expression in Eukaryotes

Regulation of gene expression in eukaryotes

Gene expression in a multi-cellular organism is very different than in a prokaryote. One such difference is cellular differentiation.

-Cellular differentiation: start out with general undefined cells then turn on different genes to make a different type of cell or cell types. This process is CRITICAL. You need to turn on the correct genes at the correct times otherwise you will get death.

• ex. of cellular differentiation --> you need different things in a muscle cell then in a nerve cell.

(3:35) How the regulation of gene expression differs in Eukaryotes

• Eukaryotic cells are larger and more complex than prokaryotes. Within this, DNA is packed into chromatin with histone proteins. Chromatin remodeling is a key step in the regulation of gene expression in a eukaryote. So if you don't have your DNA in a form that is accessible to RNA POL to copy you can essentially turn off transcription by shuttin down chromatin remodiling.
  
• Eukaryotes typically have their DNA in multiple chromosomes not one chromosome (typically seen in prokaryotic organism)
  
• Because DNA is in the nucleus and the ribosomes are at the endoplasmic reticulum (ER) in the cytosol, transcription and translation are seperated spatially and temporally (occur at different times). In a prokaryote: almost as quickly transcription starts to produce RNA, ribosomes come in, bind that RNA and start to translate that RNA. In eukaryotes these processes are seperated.
• mRNA molecules are modified prior to exiting the nucleus in a eukaryote. That modification includes splicing.
  
• (8:50) Eukaryotic mRNA molecules are more stable than prokaryotic, they have a longer half-life (amount of time they exist). Partly because Prokaryotes need to rapidly respone to changing conditions. Eukarytoes don't experience this as much.
• In eukaryotes regulation can also occur at the level of translation. You have a more stable mRNA molecule but you may alter or regulate the translation of an individual mRNA molecule because it's not being produced and degrated as quickly.
  
• While all eukaryotic cells contain complete copies of their genome (all chromosomes + DNA in all cells) different cells express different subsets of genes.
• Broadly: the process of regulation in a eukaryote is a more complex process then what is seen in a prokaryote.

Differences in regulation of gene expression	Prokayote	Eukaryote
1st difference		larger and more complex
2nd difference	DNA typically in one chromosome	DNA in multiple chromosomes
3rd difference	transcription and translation happen almost simultaneously	transcription and translation are seperated spatially and temporally
4th difference		mRNA molecules are modified prior to exiting the nucleus
5th difference	Prokaryotes need to rapidly respond to changing conditions - less stable.	more stable, longer half-life
6th difference		regulation can also occur at the level of translation
7th difference		contain complete copies of their genome in each cell

(14:45) Chromosome organization in the nucleus influences gene expression

• During interphase, the DNA found in the nucleus is in a relaxed state. However, there is stil orginization to how DNA is organized within the nucleus. This organization plays a key role in the regulation of gene expression.
• Within the nucleus, each unique chromosome exists in a chromosome territory.
• The regions between the chromosome territory's are called interchromosomal domains.
• (19:45) Within the nucleus the arrangement is as follows:
• Chromosomes with small numbers's of genes have thier chromosome territory on the outside of the nuclues.
• Chromosomes with larger numbers of genes exist in chromsome territiores towared the inside of the nucleus.
• It has been proposed that the genes being actively transcribed on a chromosome will be found toward the edge of the chromosome territory. This suggests that RNA POL lives in the interchromsomal domains. The genes need to be brought close to those areas so they can be transcribed.
  
• Once we get a chromosome in position the intition of transcription begins --> two steps.
• Chromatin Remodeling
• We need to recruit a number of factors (typically proteins) that help initiate transcription.

(27:10) Transcription Initiation

• There are three common cis acting elements in Eukaryotic transcription initiation: promoters, enhancers and silencers.
• The process requires chromatin remodeling, a number of DNA sequences and over 100 proteins. Just to initiate transcription.

• Promoters: The site where the transcription machinary binds to start transcription.
• The promoter typically facilitates a basal level of transcription
• Typically adjacent to the gene (upstream)
• contains a few 100 nucleotides.
• (31:30) Within those nucleotides there are number of key sequences of the promoter
• TATA Box a.k.a. Core promoter - 25-30 BP region of DNA that is bound by RNA POL. This contians a 7-8 BP consensus sequence. The consensus sequence contains a number of nucleotide sequences and the TATA sequence. This is the RNA POL "docking site".
• mutations in this sequence decrease the level of transcription. Deletion of the sequence results in loss of transcription.
• CAAT Box - these element contain the sequence CAAT or CCAAT located 70 BP upstream of the start of the gene.
  
• Mutations in this sequence=decreased transcription.
  
• (36:15) GC Box - GGGCGG and is found 110 BP upstream of the transcription start site.
  
• both the GC box and the CAAT box can serve as enhancers as well as part of the promoter.
  

• (40:00) Enhancers: Can be found on either side of a gene and can some times be great distances away.
  
• enhancers are typically bound by multiple proteins with a net effect of stimulation of transcription.
• (42:50) How do we differentiate an enhancer from a promoter?
• Promoter regions are found at fixed locations. Enhancers are not found in fixed locations (they can move around.
• You can invert an enhancer without affecting its activity. (not the case for a promoter)
  
• If you move an enhancer to a different gene, that gene will now be regulated by the enhancer.
  
• Promoters are responsible for the basal level of gene expression. Enhancers are necessary for the full expression of a gene.
  
• Enhancers can be cell type specific and promoters are not.

(47:45) How do enhancers stimulate the level of transcription?

• Factors can bind enhancers which help with chromatin remodeling.
  
• When a factor binds the enhancer it bends the DNA bringing the enhancer and promoter closer together. This can help stimulate RNA POL binding to the promoter
  

Genetics Lecture 30, 11/17: Chapter 16, trp Operon, Monod, Yanofsky and Bertrand

lab wed. pre lab write-up is water testing

For Powerpoint slides copy and paste this link in a new window: http://docs.google.com/Presentation?id=dhqwrndc_501cs7scpdq

1953- Monod and coworkers found an operon which is repressible. It was the trp operon.

Trp operon produces enzymes which help the cell synthesize tryptophan.

• (2:35) The presence of tryptophan turns off this operon (trp operon).
• Monod et. al. proposed that the repressor was normally inactive. It was turned on in the presence of tryptophan. This further led them to propose that the repressor had a co-repressor.
  
• What they figured out: the regulation was done with the assistance of constituitive mutants.
  
• 1st constituitive mutant was trpR- strain which produces the trpR protein that functions as a repressor protein
  
• (7:25) 2nd was a mutation in the operator region (analogous to lac O).
• In the absence of tryptophan the trpR repressor protein cannot bind the operator. Because the operator is unbound transcription takes place.
• (9:40) In the presence of tryptophan . . . the trpR repressor protein is made. Tryptophan binds to the trpR repressor. When this happens the repressor binds to the operator and prevents transcription.
  
• (12:20) 5'-UTR - an untranslated region. A sequence of DNA which is transcribed upstream of a gene but not copied. Typically they are involved in gene regulation.
  
• The untranslated region in the trp operon is 162 BP and it is transcribed but not translated. This untranslated region is studied by Yanofsky and Bertrand.

(17:15) Yanofsky and Bertrand (Slide Title: The role of the hairpin in the regulation of the trp operon).

• Irregardless of the presence or absence of tryptophan transcription of the trp operon begins.
• In the presence of tryptophan transcription ceases 140 BP into the 5' untranslated region. (140 of 162 are translated)
• In the absence of tryptophan transcription begins and continues to the end of the operon.
• There is a site between 115 and 140 nucleotides into the 5' untranslated region that serves as an attenuator sequence. (21:50) What happens is that this region can form hairpins in the 5'-UTR RNA.
  
• (23:30) In the presence of tryptophan the attenuator region forms two hairpins. This causes transcription to stop. In the absence of tryptophan the attenuator forms a single hairpin (called the anti terminator), transcription is able to complete.
• 
  

Genetics Lecture 29, 11/12; Chapter 16, Lac operons, Jacob & Monod

See the Power Point for Chapter 16 here
http://docs.google.com/Presentation?id=dhqwrndc_501cs7scpdq



Quick Review:
Last time we talked abou the lac operon and how we regulate the expression of the lac operon in the presence and absence of lactose.
Function of the lac operon is to help convert lactose to glucose and galactose.

What happens when both lactose and glucose (primary metabolite) are present?

• ANS: In the presence of glucose the lac operon is off. (fail to transcribe the lac operon because you don't need to break the lactose down)
  
• so then . . . "How is this accomplished?". It is accomplished through a process called CAP (Catabolite activator protein) and cAMP (cyclic AMP)
  
• Quick Review: RNA POL on its own has a hard time finding promoters. (That's why we have the sigma factor, it aids the RNA POL in getting to the promoter)
  
• (3:40) The CAP is always present in the cell. CAP binding to the promoter of the lac operon is necessary for RNA POL to bind a promoter. So without the binding of CAP RNA POL seldom binds to the promoter.
  
• CAP can only bind to the promoter for the lac operon when its complexed with cAMP. When cAMP is not present, CAP does not bind to promoter.
  

(7:00) cAMP levels are dependent on the level of glucose in the media.

• cAMP is a derivative of ATP
• One of the enzymes utilized in cAMP production is called adenylcyclase. What happens is that the production of adenycyclase is inhibited by glucose. Thus, when glucose is present in the media the adenylcyclase levels fall. As they fall cAMP levels fall. SUMMARY: in the presence of glucose you have low levels of cAMP, therefore, you rarely form the cAMP - CAP complex, therfore, you rarely transcribe the lac operon.
• Summary of this example: Lactose is present in this example which means lactose is able to bind to lac I bind to the repressor protein, inhibit the repressor proteins ability to bind to the operator. If RNA POL can bind, it can copy BUT the presence of glucose reslts in low cAMP and the lac operon is off.
  

(11:35 - 21:25) Schau used the following volunteers to demonstrate regulation of lac operon.
Dan - promoter
Becky - operater
Laura - structural genes
Emily - RNA polymerase
Chris - catabolyte activator protein (his hat is cAMP)
Lindsey - Lac I
Lisa - Lac I gene
Britney - Lactose
E.J. - Glucose

(15:25) Review of Monday's lecture:

• Begin with absence of lactose - no sugar at all.
• Lac I gene gives rise to Lac I protein, Lac I protein goes to operator (Lac O). RNA POL goes to promoter and attempts to copy but Lac I stops. As a result there is no transcription that happens

(16:50) Introduce a sugar - Lactose
Lac I gene gives rise to Lac I protein, Lac I protein goes to operator (Lac O). Lactose binds to Lac I and prevents it from getting to operator. RNA POL can then come in, bind to promoter and transcribe.

(18:00) Presence of lactose and CAP - cAMP complex.
Lac I gene gives rise to Lac I protein, Lac I protein goes to operator
(Lac O). Lactose binds to Lac I and prevents it from getting to
operator. CAP helps RNA POL find the promoter.

(19:10) Presence of lactose and glucose and CAP - cAMP complex.
Lac I gene gives rise to Lac I protein, Lac I protein goes to operator
(Lac O). Lactose binds to Lac I and prevents it from getting to
operator. CAP helps RNA POL find the promoter BUT glucose reduces levels of cAMP and without it you can not get RNA POL to the promoter.

side note: Lac I repressor protein (when present and functioning) binds to operator and inhibits RNA POL from transcribing genes

(23:15) 3 different processes we NEED to KNOW

• absence of sugar
• presence of lactose
• presence of lactose and glucose.

• For each of the above conditions we should know
  
• Is the operon transcribed?
• How does the regulation of the operon take place?
• In regards to regulation include talking about cAMP and CAP and lac I repressor protein

(25:40) Additional work by Jacob and Monod

• Quick Review: The understanding of how this regulation takes place relied on these two constituitive mutants: lac I- and lac O-
  
• When these above mutations were orginally identified, nobody knew if they were mutations in cis-acting elements or trans-acting factors. We (as a genetics class) know that Lac I is a trans-acting factor and Lac O is a cis-acting element. Jacob and Monod did not know this as a result they developed the "Cis-trans test".
• (28:10) "Cis-trans test", used to determine the nature of these mutations. How they did the test:
• The lac I mutant strain in the cell constiuitively expresses the lac operon.(despite presence or absence of lactose. Normally it is sensitive to lactose, however)
  
• Sooo . . . They placed a wildtype (WT) copy of the lac I gene onto a plasmid (a piece of extra chromosomal DNA). Then they introduced this plasmid into the lac I- strain. What this accomplishes: The WT copy of the lac I gene on the plasmid will produce or give rise to the WT lac I protein. It will NOT correct a mutatnt lac I DNA sequence in the cells DNA.
  
• Important: The presence of the plasmid with the lac I gene can correct a mutation in a trans-acting factor. Not in a cis-acting element.

• (33:35) What will the presence of WT lac I protein do to the lac I mutant strain?

• We would expect that this would correct the constituitive phenotype. So the lac I mutant strain with the WT copy of lac I on a plasmid, making protein, will no longer be constituitive.
• Now that we have a normal protein it will require lactose to bind to it to prevent it from binding to the operater. After performing the test the conclusion is that lac I produces a trans-acting factor - a molecule that is a protein that is produced to regulate the expression of another operon.
  
• Summary: This cell will only express the lac operon when lactose is present

(36:55) Now they want to know about lac O.

• The lac O mutant strain in the cell constiuitively expresses the lac operon
• Is lac O a cis - acting element or a trans acting factor?
• Introduce a WT copy of the lac O sequence on a plasmid, this can produce WT copies of the lac O protein. However, the problem is, lac O does not produce a protein.
  
• The presence of the plasmid does not correct the mutant lac O sequence. Therefore, the lac I repressor cannot bind to the mutant lac O sequence.
  
• The presence of the plasmid does not correct the constituitive phenotype.
  
• Thus we conclude that lac O is a cis-acting element.
• (plasmid will correct a trans-acting factor but not a cis-acting element.)
  

(43:25) Recap of Jacob and Manod "Cis-trans test"

• cis acting element - sequence of DNA that is acted upon by a trans-acting factor to regulate the expression of a gene or operon.
  
• trans acting factor - protein that regulates the expression of a gene or an operon by acting ON a cis-acting element. Binds to DNA and either turns on or off the expression of a gene.
• In the lac operon, the binding of lac I (the repressor protein) to the operator turns off transcription. When you mutate either of these elements you mutate the protein and it won't bind to the operator if you mutate the operator the normal protein can't bind - same effect. The end result is constituitive expression - we always express the lac operon.
  
• (46:50) What Jacob and Monod started this test, all they knew was that they express constituitively. They wanted to know was it trans-acting factor or cis-acting element.
  
• They take the mutant strains, started with lac I. They have an abnormal copy of the lac I gene.
  
• They go to a different cell and get a normal copy. They put this normal copy into the lac I mutant strain. It will be transcribed and you will make WT copies of the lac I protein. So instead of a cell that only has "bad" versions of lac I protein that can't bind to the operator, we have a cell that also has normal versions of the protein that can bind to the operator. When that happens the cell that was always expressing the lac operon is now only going to express the lac operon when lactose is present - so there is a change in the phenotype of the cell. BECAUSE that works for lac I we say that lac I is a trans-acting factor. If it DID NOT work we would have called it a cis-acting element.
  
• (50:00) They then repeat the process for lac O.
• They take a mutant copy of lac O and put in a normal version (from some other cell). It is a cis-acting element.
  

Genetics Lecture 28, 11/10; Chap 16: Regulation of Gene Expression, Lac operon

Test Friday 11/16

Chap 16: Regulation of Gene expression in Prokaryotes

• Gene Expression: Transcription - process of converting DNA to RNA
• The cell regulates transcription and can turn on or off the transcription of a gene. It doesn't want to make something it won't use. (sensitive to how they spend energy)
  
• In prokaryotes genes are typically found by themselves, a single gene OR as operons.
  
• Operons - a group of genes which are transcribed as a single unit.
• Typically the genes found in an operon are all involved in some process together (this way they can all be regulated together)
  
• Promoter - upstream of a gene (closer to 5 prime end) site where RNA POL binds to start transcription.
• In prokaryotes there is a sigma factor - a small protein that assists RNA POL in recognizing and binding a promoter.
• After that RNA POL begins copying DNA into RNA and continues until it reaches the Terminator (which is at the end of the gene or operon) and signals the end of transcription. No more than that is copied - don't want to waste energy.
  

(8:20) How do we regulate the above process of transcription. The study of gene regulation in prokaryotes is EXTENSIVE.

• 1900 it was recognized that cells fail to produce the enzymes for lactose metabolism when lactose is absent. Gave rise to the idea that gene expression is adaptive.
  
• Constituitive expression - a gene that is always expressed at a relatively high level.

(11:20) Two types of Expression:

• Positive Regulation - the turning on of the expression of a gene. (Gene is off and then you do something that induces the expression of that gene)
  
• Negative Regulation - the gene is being expressed until you turn it off. (ex. Tryptophan)
  

(14:15) Example of Positive Regulation: Lac Operon

• Late 1940's - Jacob + Monod (links to Wikipedia article): laid the groundwork for all the understanding of gene regulation.
  
• Based on the idea that in a prokaryote the enzymes for lactose metabolism are off without lactose. Their goal was to understand this idea.
  
• lac operon: three structural genes involved in lactose metabolism
• lac Z - codes for beta galactosidase which breaks down lactose to produce glucose and galactose. You don't want to produce this if you have no lactose.
  
• lac Y - produces an enzyme called permase which is responsible for facilitating the entry of lactose into the cell.
  
• lac A - codes for transacetylase. It is believed that it helps to breakdown some of the toxic byproducts of lactose metabolism.
• cis-acting element - a DNA sequence that is bound and acted upon to allow the regulation of a genes expression (turns it on or off).
  
• trans acting factor - a molecule, often a protein, that binds a cis-acting element to regulate the expression of the gene.
• (24:00) Gratuitous inducer (found by Jacob and Monod) - a molecule which mimics the activity of a molecule which normally activates a system.
  
• Lactose induces the expression of the lac operon. The inducer can fill the role of lactose.
• IPTG (part of the gratuitous inducer) - when added to the system turned on the expression of the lac operon. This allowed them to find constitutive mutants (mutants which always express the lac operon).
• Lac I - gene upstream of the lac operon (not part of the operon). Lac I is repressor gene that produces a protein which bound and repressed the lac operon. When you mutate the lac I gene, cells consituitively express the lac operon. (Trans acting factor - something that is produced that acts upon a cis-acting element) It no longer represses the expression of the lac operon.
  
• Lac O - mutation in the operator sequence. Cis acting element. DNA sequence bound by Lac I to prevent transcription of the lac operon. (this is a constituitive mutant)
• Finding the lac I and lac O mutants allowed them to develop a hypothesis as to how the lac operon worked
  

(33:10) How the regulation of the lac operon works in the absence of sugar.

• The lac I gene produces the Lac I protein. The repressor protein binds to lac O (operator sequence)
• When this happens RNA POL binds the promoter. Repressor protein blocks RNA POL from copying the lac operon, therefore, operon repressed.

How the regulation of the lac operon works in the presence of lactose

• lac I gene produces the Lac I repressor protein. Lactose binds to the Lac I repressor protein. This triggers a confirmation change in Lac I.
  
• Because of this Lac I cannot bind the operator. SO when this happens RNA POL can bind the promoter and copy the lac operon. The result is that the lac operon is induced.
  

Genetics Lecture 27, 11/5; Chapter 15, mutations, DNA repair

No audio available for this lecture.

How do you determine if a compound is mutagenic? - AMES Test

• 
  Use 4 strains of the bacteria strain salmonella typherium (all oxytrophic which means they can't grow on minimal media)
  
  


• 
  One strain is especially susceptable to base pair substitution.  The other 3 are susceptible to frameshift mutation 
  


• 
  (3:10) Can this compound cause mutation and result in a significant appearence of prototrophs.
  




• 
  Auxotrophic strain + liver enzymes.  Many mutagenic compounds only become mutagenic after exposure to liver enzymes.
  


• 
  Plate the strain on two minimal media plates
  


• 
  Mix your potential mutagen with liver enzyes - add to a piece of filter paper
  


• 
  filter paper is placed on one of two minimal media plates
  


• 
  Incubate overnight and determin number of prototrophs on both minimal media plates.
  




• 
  if experimental plate has a substantially larger number of prototrophs then the control then compound is a mutagen
  


• 
  If your experimental and control plates have similar numbers of prototrophs then it is most likely not a mutagen.  (test was developed in 1970's and was used on carcinogens, 80% resulted in mutation)

(13:15) DNA Repair

• Photoreactivation repair (involves prokaryotes) - requires activation from UV light
• How it was discovered:
  
• What they knew: If you exposed DNA to light from the blue spectrum shortly after exposing DNA o UV light you can reverse some damage done to the DNA.  Relied on light-dependent enzyme which was also temperature sensitive.
• What they discovered: Found a protein called photoreactivation enzyme (originally isolated from E. coli) and discovered that this eznyme absorbed 1 photon of light and thus became active.  As it became active the enzyme gains the ability to cleave the bonds in a thymine dimer.  
• Nucleotide excision repair (found in the vast majority of organisms)  3 step process.
•  - Begin with damaged piece of DNA -
  
• Nucleases search for and identify these lesions.  When they find it they cut out the damaged DNA along with a number of bases on either side.  This results in a gap in the DNA.  
• (23:40) Form of DNA polymerase which fills in the gap with new nucleotides and in the process corrects the mutation.
• (25:05) DNA ligase functions to seal gaps in the DNA strand.
  
• (26:45) Base excision repair: Corrects minor alteration to DNA molecules (a bit more focused then nucleotide excision repair)
  
• DNA glycosylase (an enzyme) functions to cleave a base from the sugar phosphate backbone
• AP endonuclease recognizes a sugar without a base (whatever glycosylase leaves behind).  When it does this is leaves a nick in the DNA strand upsteram of the sugar where there is no base.
• DNA polymerase comes in and removes the sugar and replaces it with a new nucleotide.
• DNA ligase comes and seals the gap in the DNA strand.
• (32:20) Proofreading during DNA replication
• During DNA replication, DNA polymerase functions to add nucleotides to a growing DNA chain.
  
• After DNA polymerase adds a base it pauses to check that base looking specifically for "correctness".  Did the polymerase add the right base.  TWO things can happen: if the base is correct DNA polymerase moves on.  If the base is not correct DNA polymerase (with its 3' to 5' exonuclease activity) will function to cleave and remove an incorrect base.  After removal DNA polymerase replaces the base and moves on.
• (37:10) SOS repair system: responds to gaps in DNA strands
  
• Use an alternative DNA polymerase (DNA POL V).
• Typically gaps cause DNA replication to stall out.  DNA POL V comes to a gap like this and can fill, it solves the problem.  However, in filling the gap, DNA POL V uses less stringent base pairing rules.  By-product is increased mismatched base pairs.
  

Genetics Lecture 26, 11/3; Chapter 15, Mutations, Tautomeric Shift, Spontaneous DNA changes

Observations on the fact that the rate of mutation varies from organism to organism.

• The rate of spontaneous mutation is exceedingly low
• Great deal of variety from organism to organism.
• Within a single organism different genes mutate different rates

Deleterious mutations.

• 5-10% of the DNA in an organism actually codes for protein. Majority of mutations are in non-coding regions.
• Deleterious mutation: Any mutation that causes a change in phenotype.
• On average you will see 1.6 deleterious mutations/person/generation

(6:00) Specific changes a mutation can create on a DNA sequence

• Missense mutation: a single nucleotide change in the sequence of a gene. EFFECT: Typically is that there is a single amino change (sickle-cell anemia). Also known as Point Mutation and Base substitution.
• Two types
• Transition: exchange purine for purine or pyrimidine for pyrimidine
• Transversion: exchange purine for pyrimidine.
• Frameshift mutation:
• deletion or insertion: change the reading frame (way a gene is read to be converted to protein) for the gene. Change both the sequence and the length of the protein.
• Nonsense mutation
• create a mutation which creates a stop codon where a stop codon was not present previously. Results in prematrue termination of translation.

(15:15) Chemical agents that can damage DNA

• Tautomeric Shift - alternative versions of a compound. Nitrogenous bases have tautomeres.
  
• If a nitrogenous base undergoes a tautomeric shift. The tautomere of the nitrogenous base typically experiences non-watson&crick base pairing.
• Compounds that function as base analogs: Mutagenic chemicals with the ability to substitute for one of the bases.
• (19:45) EX. 5-Bromouracil: a derivative of uracil. Functions as an analog for thymine. Normal version of 5-Bromouracil base pairs with adenine. 5- Bromouracil can undergo tautomeric shift. When this happens base pairs with guanine. The presnce of 5-Bromouracil makes DNA more susceptible to mutagenesis with UV light.
• (23:45) EX. 2-aminopurine: analog for adenine. Base pairs with cytosine.
• Alkylating agents --> EMS (Ethylmethyl sulfonate). Add an alkyl group to a base. Creates bases that act like base analogs. EMS alkylates guanine and creates 6-ethyl guanine which functions as an analog for adenine and thus binds to thymine.
  
• (29:45) Arsidine Dyes - cause frameshift mutations. They tend to remove 1 to 2 bases from a sequence of DNA in their role as an intercalating agent. Intercalating agent - a compound that is similar in size to a nitrogenous base pair. Because of this it is able to slip in between spaces in a DNA sequence. This creates torsion on the DNA and the DNA loses 1 to 2 BP in order to relieve the torsion.
• EX. of intercalating agent: Proflavin, Arsidine orange, Ethidium bromide.

(33:40)Environmental/ Spontaneous changes to DNA

• Depurination: spontaneous loss of a purine base from its sugar-phosphate backbone.
• Apurinic site - gap in the sequenc of bases. This is a problem in DNA replication
• EFFECTS:
• DNA replication STOPS.
• DNA polymerase gets to the opening and inserts any base (most likely to be wrong)
• Deamination: nitrogenous base has an amino group converted to a keto group. Cytosine is converted to uracil. The uracil then base pairs with adenine. Adenine can also undergo a deamination. Adenine will become hypoxanthine and then base pairs with cytosine.
• Thymine Dimers: occur when you have two bases of thymine next to each other on a single strand of DNA. UV (260nm wavelength) hits the DNA and the two bases of thymine break their bonds with the base that they are paired to. Instead they form a double bond with each other. This can cause large scale mutations during DNA replication to the point that you can see the death of an organism. This is why UV light can be used to sterilize things.
• 
  

Genetics Lecture 25, 10/31; Chapter 15 DNA repair and mutation

DNA repair and mutation
Looking specifically at alterations to the chemical composition of DNA

• (1:45) Mutation is the basis of genetic study. Mutate a gene and the determine the phenotype
• Certain organisms are easier to mutate, for example: bacteria, fungi, yeast, viruses and fruit flies. THey have shorter lifespans.

(4:35) How mutations arise.

• Two models for how bacterial mutations arise
• Spontaneous: mutation occurs at any given time
• Adaptive: cell will mutate in response to environmental pressure
• EX. Antibiotic resistance: the cell will mutate in response to the presence of the antibiotic.
• 1943 - Luria and Delbruck set out to test the two models. They set up a number of cultures of e. coli. Sensitive to infection with the T1 phage. Prior to starting the counted the total number of cells. They then exposed the cells to phage. Following exposure they counted number of survivors.
  
• Survivors are cells witha mutation that allowed them to survive infection.
• start with small # of cells and grow them.
  
• Spontaneous mutation could occur at any time.
• if it happens ealy 1 cell picks up a mutation and divides. All cells from this cell have the mutation. High percentage of cells with a mutation
• if it happens late you get orginal cell with the mutation and a few rounds of cell division and a small percentage of resistant cells.
• in the experiment they looked at 10 cultures. SPontaneous model would propose that there would be variety in the number of resistant cells in the cultures. In the adaptive model mutation is the result of environmental pressure. THis suggests that with a number of different cultures, if all cells are exposed to phage at the same time then each culture should have mutation occuring at the same time. Each culrue should have about the same number of reistant cells - adaptive model
• found that 10 cultures had a wide variety of resistant cultures. Indicates the mutation is spontaneous. Reffered to as the fluctuation test.
  
• Additional work has been down to show that mutation is primarily spontaneous. There are rare occasions when mutation is adaptive.

(23:00) How to classify mutations

Spontaneous vs. induced.

• Spontaneous mutation arises naturally in a spontaneous fashion. No exposure that led to the mutation.
• Induced mutation: some envionmental exposure leads to a mutation. EX: Smoke two packs a day and you develop a mutation --> lung cancer.

(27:10) Somatic mutations vs gametic mutations

• somatic mutations arise in somatic cells. A somatic cell mutation typically leads to a small population of cells with that mutation.
• Gametic mutations - mutations in gametes. A gametic mutation in a gamete involved in a fertilization, that offspring will have that mutation in every cell.

(32:20) Broad categories of mutation

• morphological mutation. A mutation which alters the structure of a all organisms.
• Nutritional mutation: Lack the ability to synthesize an amino acid vitamin or a mutation that prevents you from breaking down some compound.
• Behavioural mutation: any mutation which alters the behaviour of an organism. EX. Fruitflies - beating of wings is important in mating. A mutation that impacts the ability of a fruit fly to Beat its wings will have behavioral consequences.
• (38:00) Regulatroy mutations: mutations in genes which produce proteins that regulate the expression of other genes. Alter the way the regulated gene is expressed.
• (40:10) Lethal Mutation: any mutation which kills an organism
• Conditional mutation: only seen under certain conditions.
• EX temperature sensitive mutation: a mutation that is scene at a particular temperature.
  

Genetics Lecture 24, 10/29; Large Mutations in Chromosome Structure

Side Note:
Trisomy - 3 copies of one chromosome.
Triploidy - instead of 46 you have 69 chromosomes because you have extra copies of every chromosome

(1:00) Large mutations in Chromosome structure

Deletion: removal of a substantial (as in a whole gene) portion of a chromosome.

• Two Categories of Deletion:
  
• Terminal Deletion: delete from the end of a chromosome
• Intercalary deletion: an internal deletion within the structure of a chromosome
• Cri - du - chat syndrome: a deletion of a small portion of a chromosome results in a substantial effect (baby cries like cat)
• In any deletion, if it is large enough, it will result in death of the organism.

(6:45) Gene Duplication Mutation: a portion of a chromosome copied. Could be anywhere from a single locus to multiple locus, caused by unequal crossover during meiosis.

• Gene redundancy/amplification: deals particularly with essential genes example rRNA gene.
• An organism will have multiple copies of the rRNA gene. rRNA is represented by 0.3 to 0.4% of all genes in an organism. Drosophila has 130 copies of the rRNA gene. If you delete one you have other copies to fill this role. If only a single copy is present of an essential gene, mutation will most likely result in death. In Drosophila, a mutation which decreases the number of copies of the gene still equals poor development and reduced viability. The larger the number of mutated genes the greater the severity.
• (13:25) Gene duplication can lead to phenotypic effects for the cell. Bar eyed phenotype in Drosophila (X-linked dominant mutation)
• heterozygous females and hemizygous males (mutant X chromosome) have one version of the phenotype.
  
• homozygous females have a more severe version of the phenotype.
• in 1930's it was deteremined that all flies have one copy of a particular region of the X chromosome. Your are wild type of you have one copy. The Bar eyed phenotype has two copies of this region. The homozygous females a.k.a. Double bar eyed (the most severe) have 3 copies of this region.
  
• (19:15) Gene duplication could serve as a resovoir for new genes to be produced
• if you have a single copy of an essential gene mutation has a potentially devastating effect.
• if yoy have MULTIPLE copies of a gene you can make one copy and make a new gene with a new function and still have copies of the original gene.
• Evidence that supports this:
• If you compare genes with very different functions there is a relatively high degree of sequence similarity.
• Families of genes. Groups of genes within an organism which are highly similar (80-90% similar) and share similar functions. EX. aminoacyl tRNA synthases: enzymes which link amino acids to tRNA molecules. EACH cell has 20 different forms of this enzyme 1 per unique amino acid.

(26:30) Inversion: a region of chromosome flipped 180 degrees, reordering the genes

• pericentric inversion: the centerosome region is included in the region
• paracentric inversion: the centrosome is not involved
• Two effects of inversion:
• an in version can fuction to prevent synapsis during meiosis
• position effect: being near a region of heterochromatin can change the expression of a gene. If the inversion does this (changes the position relative to heterochromatin) it can be expressed.

(30:50) Translocation: a region of a chromosome moving to an entirely different region of the genome

• major impact is position effect: moving genes closer or farther away from regions of heterochromatin.
  

Genetics Lecture 23, 10/27 - Chapter 8: Chromosomal mutations, non disjunction, monosomy, trisomy

Chromosome Mutation: anytime you have a cell with an abnormal # of chromosomes. Typically this is the loss or gain of one chromosome.

• Monosomy: loss of one chromosome
• Trisomy: gain of a chromosome . . . 3 copies of a chromosome
• QUES: How does this happen? ANS: non-disjunction event during meiosis.
  
• Non-disjunction event explained: during anaphase when a chromosome pair is supposed to separate it fails to do so. Because of this you produce gametes which are not haploid. Your gametes are either diploid or empty for that particular chromosome.
• When you fertilize a diploid gamete you produce a trisomy. If a gamete without a copy of a chromosome is fertilized you will get a monosomy.
• Anueploidy - a condition of having one extra or one less copy of a single chromosome.
• Euploidy - condition where a cell has a complete extra haploid complement of chromosomes. EVERY CHROMOSOME
• Triploid - has 3 complements of chromosomes
• tetraploid - 4 haploid complements of chromosomes
• Result - broad phenotypic changes.

(13:05) Genetic mutations more in depth

• Monosomy: a condition of having one copy of a particular chromosome.
  
• Partial Monosomy: Cri-du-chat syndrome (cry of the cat)--> deletion of a large portion of the P arm on chromsome 5. Referred to as 46, -5p
• Anatomic abnormalities in the cardiac region as well as the gastrointestinal tract and mental retardation.
  
• Abnormal development of the glotis and the larynx. When these baby's cry they sound like cats.
• There is a correlation between the size of the deletion and the severity of the disease. The larger the deletion the worse the disease.
• (20:00) Trisomy: (more prevelant/common) presence of an extra copy.
• Trisomy 21 = Down Syndrome.
• series of syptoms associated with down syndrome
• affected individual - 6-8 of the traits
  
• affected individuals have a tendancy to look like each other.
• Traits: epicanthic fold over the corner of the eye. Flat face. Round head. Short. Some form of mental retardation. Short life span (avg into 50's)
• Typically the non-disjunction is in the ovum (95%). Correlation between occurency and the age of the mother at conception.
• If mom is 30 it is 1 in 1000.
• If mom is 40 it is 1 in 100
• If mom is 45 it is 1 in 50
• 35% of all births to a Mom above 40 result in Down Syndrome.
• (28:20) Why is there such a strong correlation between age of the mother and frequency of Down Syndrome?
• Females are born with all of their eggs - the eggs are with the mother since birth. Meiosis II happens during menstrul cycle and fertilization.
  
• (32:55) Trisomy 13 - Patau syndrome
• Infant is not mentally alert. High chance of being deaf. Clef Palate.
  
• Typically death occurs by 3 months.
• Upon autopsy you see a high degree of organ system malformation.
  
• Average age of the parents of Patau syndrome is 32.
• (36:45) Trisomy 18 - Edwards Syndrome
• smaller than average, elongated skull, congenital dislocation of the hip.
• Life expectancy is 4 months
• majority of affected infants are female.
• average age of parents of Edwards syndrome is 34.7

• Technically you can have a trisomy or a monosomy of any chromosome. It depends on which chromosome has a non-disjunction event. Embryos with these arrangments lead to early miscarrage.
• 15-20% of all conceptions end in miscarriage. Could be as high as 50% (mother doesn't even know she was pregnant)
• Of miscarriages, 30% have a chromosomal abnormality.
• 6% of all conceptions have a chromosomal abnormality.

• Post implantation (refers to time when it occurs). What happens: there is a system set up in which the embryo is, in effect, screened for abnormalities. If an abnormality is found your body will attempt to trigger a miscarriage. WHY? Basic answer is, conserve energy. Historically, there was a high mortality rate for women during birth - body wants to preserve itself.
  

• Miscarriages are on the rise. WHY?
• People are delaying having children till later in life.
• Home pregnancy tests - people are finding out they are pregnant sooner. (Schau's Theory)
  

Genetics Lecture 22, 10/22 - Eukaryotic Chromosome Structure, Problems with Packing DNA

Eukaryotic Chromosome Structure

• Dependent on the cell cycle
• Interphase --> chromatin --> less condensed form of DNA
• mitosis --> chromosome 10,000 fold condensation of chromatin to form distinct structures.
  
• overall organization of a eukaryotic chromosome is greater than the organization of the prokaryotic chromosome.
• E. Coli has a single chromosome with a length of 1,200um. Humans have multiple chromosomes (46) and the length is 19,000 to 73,000um.
• Two protiens that associate with DNA during condensation
• histones - major protein responsible for packing
• overall charge on a histone is positive. DNA has a negative charge.
• non histone proteins

(6:10) Evidence for how the packing of DNA in eukaryotic chromsomes occurs

• 1.) digestion of chromatin with an endonuclease. Yields fragments that are consistently 200bp in length. Suggests that the DNA in chromatin is in a repeating unit
  
• 2.) microscopic visualization shows that chromatin fibers are linear arrays of spherical particles. Look like "beads on a string"
• 3.) observed that histone proteins could interact with each other. In this interaction that formed a tetromere structure and proposed that one tetromere associated with 200 BP of DNA.
• 4.) refined endonuclease digestion data. 146 BP of DNA that associates with two tetromer of histones.

PACKING

• (12:30) First level of chromatin packing. Nucleosom core particle or the 11nm fber. 146 BP of DNA wrap around two tetromere of histones. End result is DNA with a diameter of 11nm.
• second level. 30nm fiber. Solenoid structure - you have a number of nucleosome core particles. These align around a central histone called the H1 histone. Creates a DNA molecule with a diameter of 30nm, chromatin exists as teh 30nm fiber.
• Mitotic chromosome with a diameter of 1400nm

(19:45) PROBLEMS WITH PACKING DNA

• packing leaves the DNA inaccessible to certain non-histone proteins. Ways to get around this problem is called Chromatin remodiling.
• (22:45) Chromatin remodeling: allows the packing of DNA to be temporarily relaxed so it can be replicated or transcribed.
• this is done by modifying amino acids on the histone proteins to weaken their association with DNA:
  
• Acetylation: add an acetyl group to lysine. this removes the positive charge on lysine.
• methylation: add a methyl group to lysine and arginine thus altering charges to allow brief access
• phosphorilation: add a phosphate to either serine or histidine. END RESULT: weaken the association between the histone and the DNA.
  
• (28:40) Within a chromosome there are different regions - two different levels of DNA packing
• Euchromatin - DNA that is undergoing normal packing. The genes in those regions can be expresed. APX. 90% of the DNA in the cell.
  
• Heterochromatin - DNA that has undergone extreme levels of packing. Highly condensed in comparison to euchromatin. Because of this the genes in these regions are not expressed.
• Unique features of heterochromatin:
• 1.) regions of DNA that are heterochromatin are gentically inactive.
• 2.) regions of heterochromatin are replicated later in the cell cycle (S phase). It is proposed that heterochromatin is important to the structural integrity of the DNA.
  
• 3.) Heterochromatin is unique to Eukaryotes.
• (34:55) What's composed of heterochromatin
• centromere, telomeres - sequences on the end of linear chromosomes, the majority of the Y chromosome
• Often see that regions adjacent to heterochromatin exhibit position effect. Genes in these regions are not expressed.

(38:35) REPETITIVE SEQUENCES ON CHROMOSOMES

• centromere - location of chromsome attachment during mitosis and meiosis, spindle fiber attachement. Within this there is a CEN sequence which is composed of three parts.
  
• 1st and 3rd - regions that are found on all chromosomes with high similarity.
• 2nd region is unique from chromosome to chromosome. BUT they're similar in homologous chromosomes.
• (42:20) Did mutational analysis on these sequences --> found that the 3rd is most critical to the function of the centromere
• (43:30) Telomere - repetitive sequence found on the end of linear chromosomes. Contains a number of repeats of the sequence GGGATT. Telomeres vary in length. All the individuals have different lenghts of telomeres. Role of telomere: protect the DNA at the end of teh chromosome from degredation. Only 5-10% of the DNA in an organism is usde in genes.

47:20 - whats on the test for 10/24

Genetics Lecture 21, 10/20 - DNA organization into chromosomes,

DNA organization into chromosomes

• Viral and bacterial chromosomes relativley simple in comparison to eukaryotic chromosome
• typically we have a single piece of nucleic acid without associated proteins. By comparison eukaryotes typically have multiple chromsomes complexed with a number of proteins.
  

• Viral chromosomes: come in variety --> DNA or RNA and these can be single stranded or double stranded, furthermore they can be linear or circular.
• Size range of viral genome: 2-52 micrometers in length. The viral genetic material undergoes packing to fit into the phage head.
• (6:00) Bacterial Chromosomes - always double stranded DNA. The DNA lives in the nucleoid region (an area in the bacterial cell where the DNA congregates). No membrane surrounding the nucleus of the DNA.
  
• (8:35) E. Coli
• Single circular chromosomes - 1.2 mm in length. It is associated with a few DNA binding proteins. Circular chromosomes have the ability to supercoil. How was this figured out?: 1965 they took viral DNA from a mouse and did a density gradient centrifugation - Take something we want to seperate based on size and put it in a tube and the lighter the molecule the faster it travels. Upon doing this they found 3 distinct species of DNA in the centrifuge. One moved slowly and thus it was believed to be linear DNA. The other two moved more quickly. One of the factions was underwound circular DNA and the other was relaxed circular DNA.
• (17:00) Relaxed circular DNA - has the standard number of twists for that pieces of DNA. Circular DNA can be underwound.
• (18:10) Underwound DNA - DNA is unwound (fewer twists) and creates stress on DNA molecule. WHen this happens it allows for the DNA to supercoil. This supercoil relieves the stress the DNA feels of being unwound. Overwound DNA has the same effect in that it causes stress and results in supercoiling.
• (21:10) How we determine when and if supercoiling will occur?
• Linking # --> predicted # of turns for a DNA molecule. Dividing total # of BP = 10.4 # of BP/turn
  
• EXAMPLE: SV40 virus - 5200 BP in its genome
• 5200/10.4 = 500 (predicted)
• Actual linking number = 475
• To determine the number of supercoils you take the actual linking # and subtract the predicted linking #. This gives us 25 negative supercoils in our DNA
• Topoisomers - two DNA molecules of the same DNA with two different structures.
• (27:25) Topoisomerase - Enzyme responsible for converting DNA from one topoisomerase to the next.
  
• (28:50) Specialized Chromosome Structures
• Polytene chromosome: very large chromosomes readily viewed under a microscope. Found in the cells of a number of tissues. They are seen as a series of alternating bands (DNA) and interband regions.
• (31:40) How do we get polytene chromosmes?
• These chromosomes are always found as homologous chromosomes pairs. Typically chromosomes only pair during mitosis and meiosis.
  
• During DNA replication, these chromsomes replicate. But the DNA is not dispersed into new chromosome pairs/new cells. A chromosome with 5000 strands of DNA stacked together puffing occrus when teh DNA strands seperate to allow activity to occur
• (35:50) Lampbrush chromosome - received its name because it looks like a lampbrush (used to clean kerosene lamps in the 19th century)
• found in the oocytes of sharks and a number of vertebrates. Also in spermadocytes in insects.
  
• It is found during prophase 1
  
• Lampbrush chromosome functions to direct metabolic activities during meiosis I