Announcements A reading assignment on DNA hybridization has
Announcements -A reading assignment on DNA hybridization has been posted on the course website. -The second midterm exam (Monday, May 19 th) will cover material through today’s lecture and next weeks quiz section. -My office hours: Fridays, by appointment between 1: 00 & 2: 30 PM. -Emailing questions to course TA’s…
If you don’t have much background with molecular biology techniques: -The textbooks that are on reserve in the undergraduate library cover these techniques very well: -Chapter 20 (specifically, pages 715 -735) in Griffiths, et al. -Chapter 9 (specifically, pages 301 -335) in Hartwell, et al. -Chapter 19 in Klug, et al. -I have provided links on the course website that direct you to sites that provide information on gel electrophoresis, Southern blotting and northern blotting. -I have placed a file on the course website that describes gel electrophoresis and Southern blotting. -The TA’s can also help-but please, not by email.
Today… -DNA cloning -DNA hybridization -Sequencing -Polymerase chain reaction
The Yeast Adenine Biosynthetic Pathway A B C D E F Y G H I J K
Given that many different genes are involved in adenine biosynthesis, what do all of these enzymes “look” like? --how are they different? --what is the sequence of amino acids? --what is their 3 -D structure? --how do the enzymes work? --do humans have the same enzymes as yeast? Bigger question: How can we figure out what any gene actually encodes? How can we get our hands on the piece of DNA so that we can sequence it and determine the amino acid sequence of the protein?
DNA cloning Dolly Clone: cells or organisms that are genetically identical because they are related by non-sexual reproduction—i. e. , by mitosis not meiosis. Copy Cat DNA clone: identical copies yeast of a DNA fragment, usually generated by propagation in bacteria Each colony is >106 identical cells.
DNA cloning—the idea A tiny fraction of a genome: ADE 2 1. Break the genome into gene-sized pieces ADE 2 2. Introduce DNA into E. coli 3. Culture each bacterium separately to generate clones 4. Study each gene in isolation Problems to solve: » How can the new DNA survive and replicate in E. coli? » How can we “select” for bacteria with the new DNA? » How do we find the one bacterium with the gene we want to study--the one with the ADE 2 gene?
Plasmids as vectors for transformation Non-essential DNA molecules that can replicate independently of the chromosome because they have an origin of replication Selectable marker Amp. R or Tet. R ori A gene that confers resistance to a drug such as ampicillin or tetracycline. For cloning: “recombine” the foreign DNA into a plasmid. In principal, vectors solve two of the problems—maintenance & selection. But how does cloning work in practice?
Restriction endonucleases Recognize and cut specific sequences in DNA Restriction enzyme Eco. RI: Eco. RI site in DNA: --GACTGAATTCTGATC---CTGACTTAAGACTAG-Reversed by DNA ligase --GACTG --CTGACTTAA AATTCTGATC-GACTAG-- sticky ends
Using Eco. RI to make recombinant DNA (Watson & Crick strands shown) partial digest with Eco. RI cut with Eco. RI --GACTG --CTGACTTAA AATTCTGATC-GACTAG-- sticky ends
Making a DNA library Tetracycline-sensitive E. coli DNA ligase to repair phosphodiester backbone transformation (Ca. Cl 2, heat shock) This is a clone tetracycline plate
A DNA library Each clone may have a different piece of the“foreign” genome in it in the form of a recombinant DNA molecule “Genomic DNA library”: • start with genomic DNA • expect all the chromosomal DNA to be represented • expect about equal representation of all sequences
Practice question The yeast genome is ~12, 000 kb and the ADE 2 gene is ~2 kb. 1. If the average insert size in the yeast genomic library is 12 kb, what fraction of the E. coli colonies are expected 1/1000 to contain the ADE 2 gene? ______ 2. If you looked at 10, 000 colonies how many are 10 expected to contain the ADE 2 gene? ______ 3. At 200 colonies per plate, how many plates would be needed to ensure that you have 10 different clones with the 50 ADE 2 gene? ______ This principle is called “coverage”--the number of times, on average, that the genome (or particular gene) is represented in the library
Quiz Section this week and next: Genetic Analysis in Caenorhabditis elegans Goals: to illustrate the logic of genetic analysis
Mutant analysis (AKA Genetic Analysis) The use of mutants to understand how a biological process normally works • Start with “unknown” system (e. g. , metabolic Locomotion pathway, embryonic development, behavior, etc. ) • Generate mutations that affect the “unknown” unc-93, unc 29, The etc. system (i. e. , athat “break” the “unknown” Is the mutation loss-of-function or gain-of-function? system) answer influences the interpretation of the phenotype Paralyzed? • Study the mutant phenotypes to reveal the hyperactive? functions of the genes etc. • Map the genes • Identify the genes (more on this later) K+-channel, Ach Receptor, etc
Loss-of-function mutations are usually recessive Amount of protein activity threshold WT phenotype mutant phenotype AA Aa aa Loss-of-function mutation simply decreases the activity of the protein; threshold determines whether dominant or not.
Gain-of-function mutations are usually dominant threshold Amount of protein activity mutant phenotype wt phenotype aa Aa AA One type of gain-of-function mutation simply increases the activity of the protein; threshold determines whether dominant or not.
Example from QS - acetylcholine receptor (unc-29) acetylcholine + outside Wild-type UNC-29 closed open inside LOF - reduced gene function, GOF - increased gene function. Phenotypes are often opposite. What would be the phenotype of a LOF mutation? - receptor missing or defective (equivalent to being always closed) What would be the phenotype of a GOF mutation? - receptor activated in absence of Ach (equivalent to being always open)
A real ACh receptor (top view) pentamer with four types of subunit: a 2, b, d, g All four types of subunits (a, b, d, g) are required for function. What happens if any of them is knocked-out by mutation (LOF)? What (probably) happens if one has GOF and another has LOF?
Example from QS: potassium channel gene unc-93 mostly closed unc-93 is expressed in muscle (body and egg-laying) outside Wild-type UNC-93 inside gate K+ outside unc-93 (cb 1500) mutant How could you make new LOF mutations in unc-93? Excess K+ flow prevents muscle excitation inside K+ unc-93 is featured again in QS 7.
Cloning by complementation! Finding the yeast ADE 2 gene in an E. coli library by complementation of an E. coli adenine mutant phenotype. Hypothesis: if the intermediates in these pathways are the same, then the enzymes should be interchangeable. E. coli pur. E+ gene HC N CH C H 2 N N R (AIR) * Yeast ADE 2 gene -OOC C N CH C H 2 N N R (CAIR) *Both species need the same basic enzyme for this step of adenine synthesis.
Cloning a gene based on its function “cloning by complementation” » Start with (recessive) E. coli pur. E- mutant strain pur. E- » Transform with DNA library made from (dominant) ADE 2 wild-type yeast strain » Ask: which plasmid clone rescues the mutant phenotype?
portion of yeast genome Library ADE 2 pur. E- E. coli no growth pur. E- no growth -adenine plate
portion of yeast genome ADE 2 must be on this plasmid! pur. E- growth! -adenine plate
Practice Questions Does that colony really have an ADE 2 plasmid? Could it not be a pur. E+ revertant or a contaminant? 1. What phenotype would the E. coli have if the plasmid was lost? (How to lose a plasmid? How exactly would you do the experiment? ) back to being ade- (unable to grow on -adenine plates)! 2. What would they expect if the original yeast genomic DNA library had been made from ade 2 mutant yeast? no colonies on the -adenine plate 3. Why is this type of functional cloning considered an example of “complementation”? 4. Can this experiment be done using the human “ADE 2” gene?
The cell division cycle Cell cycle regulation— MUCH of what we know comes from genetic analysis using temperaturesensitive yeast mutants restrictive temperature Can complement some yeast mitotic defects with human DNA! Lee Hartwell 2001 Nobel Prize
Working with DNA What else might we want to know about the ADE 2 gene? What other questions could we ask? Wesequence need to know the --what is the of amino sequence of the ADE 2 acids? gene 3 -D structure? --what is their --how do the enzymes work? --how do the mutants differ from WT? But first, a little review. .
P S S P P P S S S S S P P S S P P S S Review: DNA Structure S S P Nucleotide: repeating unit nitrogenous base P phosphate group P OH O S P N O O P P S S S P P S S denaturation breaks these H-bonds restriction enzymes break these covalent bonds H C 2 4’ 5’ N O C H H H C C 3’ OH H C 1’ H 2’ deoxyribose sugar
Review: DNA synthesis DNA polymerase requires. . . template strand 3' 5' A A T G C G A C A G T A G A T T A C G C T G T C A T C T OH OH 3’ OH primer activated nucleotides 5' 5' G OH 5' 5' C OH T 5' OH A to make the nascent strand OH
DNA denaturation (and renaturation) The strands of DNA can come apart. . . and go back together!. . TAGAGTTCTGTACGTTCGTGATGCCTGGAAC. . ATCTCAAGACATGCAAGCACTACGGACCTTG. . Heat to 95˚ (or Na. OH) Cool to ~65˚C (+ some salt) 5’. . TAGAGTTCTGTACGTTCGTGATGCCTGGAAC. . + . . ATCTCAAGACATGCAAGCACTACGGACCTTG. . 5’ It’s reversible!
DNA denaturation (and renaturation) In a complex mixture of different DNA molecules, complementary ss. DNA strands can find each other and reform ds. DNA. denaturation Na. OH or >95˚C 65˚C + salt renaturation base pairing provides the specificity to reform ds. DNA
DNA (or RNA) hybridization Annealing single strands from different sources. . TAGAGTTCTGTACGTTCGTGATGCCTGGAAC. . ATCTCAAGACATGCAAGCACTACGGACCTTG. . heat 5’. . TAGAGTTCTGTACGTTCGTGATGCCTGGAAC. . + . . ATCTCAAGACATGCAAGCACTACGGACCTTG. . 5’ cool 5’GTACGTTCGTGAT. . ATCTCAAGACATGCAAGCACTACGGACCTTG. . 5’ a “hybrid” DNA molecule. Add excess oligonucleotides with a sequence complementary to one of the strands
Some notes on hybridization -Probe and target need not be the same length both okay -Small mismatches are tolerated if overall match length is substantial okay -Small probes (25 -30 bases) can work if conditions (salt, temperature) are adjusted. Mismatches much more significant for small probes.
DNA sequencing Uses DNA polymerase The sequence obtained is for the strand being synthesized Can determine ~600 -800 bases in one “read” With 3, 000, 000 bp in the human genome, that’s a lot of sequencing reactions! Fred Sanger Nobel Prize for protein sequencing (1958) and for DNA sequencing (1980)!
Addition of a nucleotide to a 3’OH new nucleotide at 3’ end activated nucleotide H 2 O 2 +
Reagents for Sanger DNA sequencing • DNA fragment to be sequenced heat • 1 primer: short ss. DNA complementary to ONE region of the template DNA • d. NTPs All 4: d. ATP, d. CTP, d. GTP, d. TTP • DNA polymerase from E. coli Question: How does synthesizing a strand of DNA give us information on its sequence of bases? ? ? These are • small amount of each dideoxy. NTP the key
• Split the mixture among four tubes C A G T • Add a small amount of 1 dideoxy nucleotide to each reaction What’s a dideoxy? dye molecule H H A = green T = red C = blue G = black No 3’ OH. “chain terminator” • Incubate until all synthesis is complete • Analyze newly synthesized strands on a gel
Products of the four synthesis reactions “C” reaction Primer + A G A A >>>>> A GA A TTC 5 Primer “A” reaction A G A A >>>>> AGAAAA A GA A A A TTCGTA A 1 2 8 9
“G” reaction “T” reaction Primer + A G A A >>>>> A GA A TTCG 6 A GA A TTCGTA AG 10 A G A A >>>>> AGAAAAT 3 A GA A TT 4 A GA A TTCGT 7
Analysis of the new strands on a gel dd. A dd. C dd. G dd. T +etc +11 +10 +9 +8 +7 +6 +5 +4 +3 +2 +1 primer - mix all 4 reactions 3’ A T G + G A C T T A 5’-A Read bottom up
Laser scan of a sequencing gel ss. DNA fragments run past a laser where the color is detected.
Sequencing our clone The vector sequence is known; insert is to be sequenced use vector sequence for initial sequencing primers… Ampr ori end primer ori vector Amp end primer insert + insert ADE 2 gene then use the new sequence info to extend the sequencing
What if. . . you wanted to know what was different about the sequence of the ADE 2 gene from an ade 2 mutant strain? ori amp. R vector insert Do we need to start with a library from an ade 2 mutant? ADE 2 gene Not necessarily; if we already know the sequence of ADE 2 we can use PCR to obtain DNA for the ade 2 mutant allele.
The polymerase chain reaction Uses DNA polymerase Makes unlimited quantities of the DNA of interest Only requires a single template molecule—very sensitive Uses two DNA primers
Reagents for PCR 1. target DNA genomic DNA 2. Two specific primers— AAGT CACG “ left” and “right” short (17 -20 nt) oligonucleotides that are complementary to sequences on either side of the region of interest 3. All 4 d. NTPs d. ATP, d. CTP, d. GTP, d. TTP 4. DNA polymerase from the heat-loving bacterium, Thermus aquaticus doesn’t die at 95°C!
DNA products at successive cycles of PCR starting ds. DNA denature anneal synthesize cycle 1 cycle 3 repeat cycle 2
cycle 4 After 4 cycles, half of the products are DNA fragments of a specific size—the size of the DNA that lies between the two primers. By 30 cycles there would be 228 of the DNA molecules from the initial DNA molecule
Primers for PCR …must “point inward” to the region to be amplified 5’ 3’ “forward primer” 5’ 3’ region to be amplified 3’ 5’ “reverse primer”
Primers for PCR …must “point inward” to the region to be amplified 5’ 3’ region to be amplified 5’ 3’ 3’ 5’ AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGT TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCAC
Primers for PCR …must “point inward” to the region to be amplified 5’ 3’ region to be amplified 5’ 3’ 3’ 5’ AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGT TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCAC
Primers for PCR …must “point inward” to the region to be amplified 5’ 3’ region to be amplified 5’ 3’ 3’ 5’ AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGT TTGCATAAAATTGCTAAAGCTATAGGTCAAATTA AGGCTCACAT 5’ 5’AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGTA TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCAC
Practice question DNA sequence from exon 1 of the huntingtin gene 5’ GTCCCTCAAGTCCTTCCAGCAGCTTGAGCCGCCAC CGC CAGGGAGTTCAGGAAGGTCGTCGTTGTCGGCGGT What 6 nucleotides from the gray boxes could be used GGCG A. as primers to amplify the region of the genome shown above? CCCTCA “left (forward) primer”? 5’ ____3’ GTGGCG “right (reverse) primer”? 5’ ____3’ B. What size will the DNA fragment be after PCR amplification using the two primers above? 33 base pairs
C. Huntington’s disease is a dominant autosomal trait that is fatal. It usually strikes in middle age, by which time affected individuals may have already passed on the disease allele. Individuals have different numbers of the triplet CAGs in the first exon in the “huntingtin” gene. 5’ GTCCCTCAAGTCCTTCCAGCAGCTTGAGCCGCCAC CGC CAGGGAGTTCAGGAAGGTCGTCGAACTCGGCGG TGGCG How many CAG repeats are present in this allele? 4 If the CAG were in the coding portion of exon 1, how would this sequence appear in the huntingtin protein? …S L K S F Q Q L E P P P…
D. Below is a gel in which the PCR fragments of different individuals are shown. The region amplified used the same primers as in part A. Why do people have 2 bands? Because each individual represented is heterozygous for a huntingtin gene with a different number of CAG repeats Triplet repeat number 65 50 35 20 5 no disease E. What is the correlation between the number of repeats and the presence of disease? individuals with a single Huntingtin allele of more than ~ 35 CAG repeats display the disease phenotype.
F. Examine the pedigree below and determine the probability of disease among the 3 females in generation III. Notice that the middle branch of the family refused DNA testing. Numbers (i. e. 7, 21) refer to the number of CAG repeats. 45 years old 40 7, 21 7, 46 11, 15 46, 11 36 5 5 14, 46 12, 12 26 12, 46 7, 26 5 12, 26 G. Discussion questions: Would you want test results if you were II-5 or II-6? Results on the previous page indicate that individuals with a single Huntingtin allele of more than ~ 35 CAG repeats will display the disease phenotype. Individual II-5 will have this disease, whereas individual II-6 will not. Would you want your child, III-3, tested if you weren’t willing to be tested? What are the different concerns of each individual? These are personal choices, and the answers given for the other questions should provide sufficient information for you to consider these issues.
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