Molecular Basis Of Inheritance Timothy G Standish Ph
Molecular Basis Of Inheritance Timothy G. Standish, Ph. D. © 1999 Timothy G. Standish
1999 Ig Nobel Prizes � � � Science Education. The Kansas Board of Education and the Colorado State Board of Education, for mandating that children should not believe in Darwin's theory of evolution. Medicine. Arvid Vatle of Norway, for carefully collecting, classifying, and contemplating which kinds of containers his patients chose when submitting urine samples. Chemistry. Takeshi Makino, of the Safety Detective Agency in Japan, for his involvement with an infidelity detection spray that wives can apply to their husbands' underwear. Biology. Paul Bosland, of the New Mexico State University's Chile Pepper Institute, for breeding a spiceless jalapeno chile pepper. Managed Health Care. The late George and Charlotte Blonsky of New York City and San Jose, California, for inventing a childbirth device in which the woman is strapped to a circular table that is then rotated at high speed. © 1999 Timothy G. Standish
Transformation Of Bacteria Two Strains Of Streptococcus Rough Strain (Harmless) Capsules Smooth Strain (Virulent) © 1999 Timothy G. Standish
Transformation Of Bacteria The Griffith Experiment OUCH! + Control - Control Experimental © 1999 Timothy G. Standish
� The Hershey-Chase Experiement Hershey-Chase experiment showed definitively that DNA is the genetic material � Hershey and Chase took advantage of the fact that T 2 phage is made of only two things: Protein and DNA OH H H 2 N C C CH 2 S CH 3 H O H 2 N C C OH Methionine CH 2 SH O HO P NH 2 O O OH Cysteine Some amino acids contain sulfur, thus proteins contain sulfur, but not phosphorous. OH H Nucleotides contain phosphorous, thus DNA contains phosphorous, but not sulfur. © 1999 Timothy G. Standish
S 35 T 2 grown in containing media incorporate S 35 into their proteins Using S 35 Bacteria grown in T 2 attach to bacteria and inject genetic material normal nonradioactive media When centrifuged, phage protein coats remain in the supernatant while bacteria form a pellet The supernatant is radioactive, but the pellet is not. Blending causes phage protein coat to fall off Did protein enter the bacteria? Is protein the genetic material?
P 32 T 2 grown in containing media incorporate P 32 into their DNA Using P 32 Bacteria grown in T 2 attach to bacteria and inject genetic material normal nonradioactive media When centrifuged, phage protein coats remain in the supernatant while bacteria form a pellet The pellet is radioactive, but the supernatant is not. Blending causes phage protein coat to fall off Did DNA enter the bacteria? Is DNA the genetic material?
When DNA Replication Occurs � Typically DNA replication only occurs when cells are preparing to divide (there are some exceptions) � The cell life cycle is well defined and can be divided into four stages: – Gap 1 (G 1) - The growth phase in which most cells are found most of the time – Synthesis (S) - During which new DNA is synthesized – Gap 2 (G 2) - The period during which no transcription or translation occurs and final preparations for division are made – Mitosis - Cell division © 1999 Timothy G. Standish
The Cell Life Cycle Gap 1 - Doubling of cell size. Regular cellular activities. Transcription and translation etc. Synthesis of DNA Regular cell activities cease and a copy of all nuclear DNA is made S G 1 G 2 M Gap 2 - Final preparation for division Mitosis - Cell division © 1999 Timothy G. Standish
DNA Replication: How We Know � There are three ways in which DNA could be replicated: Conservative - The old double-stranded DNA serves as a template for two new strands that are then joined together, thus giving two old strands together and two new strands together ld O ld w Ne + Dispersive - In which sections of the old strands are dispersed in the new strands O Semi-conservative In which old strands serve as templates for new strands resulting in double-stranded DNA made up of an old and new strand O ld ld New O O + + + ld New O O + © 1999 Timothy G. Standish
The Meselson-Stahl Experiment � The Meselson-Stahl experiment demonstrated HO that replication is semiconservative � This experiment took advantage of the fact that nucleotide bases contain nitrogen � Thus DNA contains nitrogen � The most common form of nitrogen is N 14 with 7 protons and 7 neutrons � N 15 is called “heavy nitrogen” as it has 8 neutrons thus increasing its mass by 1 atomic mass unit OH P NH 2 O N OH N N H © 1999 Timothy G. Standish
The Meselson-Stahl Experimenttive Transfer to normal media N 14 Bacteria grown in N 15 media for several replications The conservative and dispersive models make predictions that do not come true thus, by deduction, the semiconservative model must be true. od n va ion e o r m v ti icti e se ict a v n i v d d After 20 min. rs ion er pre -co pre e i s n el (1 replication) em el sp ict o i C od D red transfer DNA S mod m p to centrifuge tube and centrifuge el X Prediction after 2 or more replications X X © 1999 Timothy G. Standish
Stages of Replicaton � Replication can be divided into three stages: � Initiation - When DNA is initially split into two strands and polymerization of new DNA is started � Elongation - When DNA is polymerized � Termination - When the new strands of DNA are completed and some finishing touches may be put on the DNA � Both elongation and termination may involve some proofreading of the DNA to ensure that mutations are not incorporated into the newly formed strands of DNA © 1999 Timothy G. Standish
Tools of Replication � There are three major enzymes involved in replication: � DNA Polymerase - Matches the correct nucleotides then joins adjacent nucleotides to each other � Primase - Provides an RNA primer to start polymerization � Ligase - Joins adjacent DNA strands together (fixes “nicks”) © 1999 Timothy G. Standish
More Tools of Replication � Helicase - Unwinds the DNA and melts it � Single-Strand Binding Proteins - Keep the DNA single stranded after it has been melted by helicase © 1999 Timothy G. Standish
Initiation � Initiation starts at specific DNA sequences called origins (Ori C = origin in E. coli chromosomes) � Large linear chromosomes may have many origins � First the origin melts (splits into two single strands of DNA) � Next primers are added � Finally DNA polymerase recognizes the primers and starts to polymerize DNA 5’ to 3’ away from the primers © 1999 Timothy G. Standish
Initiation - Forming the Replication Eye Origin of Replication 5’ 3’ 3’ 5’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ © 1999 Timothy G. Standish
Large Linear Chromosomes Have Many Origins Of Replication Origins of Replication 5’ 3’ 3’ 5’ 5’ 3’ 3’ 5’ © 1999 Timothy G. Standish
Extension - The Replication Fork 5’ 3’ 3’ 5’ 5’ 3’ 5’ Primase - Makes RNA primers Lagging Strand Okazaki fragment 5’ RNA Primers 3’ 5’ Single-strand binding proteins Prevent DNA from reannealing DNA Polymerase 5’ 3’ Helicase Melts DNA Leading Strand 5’ 3’ © 1999 Timothy G. Standish
Extension - Okazaki Fragments 5’ 3’ Okazaki Fragment DNA Pol. 3’ 5’ RNA Primer DNA Polymerase has 5’ to 3’ exonuclease activity. When it sees an RNA/DNA hybrid, it chops out the RNA and some DNA in the 5’ to 3’ direction. 5’ 3’ DNA Pol. RNA and DNA Fragments 3’ 5’ RNA Primer DNA Polymerase falls off leaving a nick. 5’ 3’ Ligase 3’ 5’ RNA Primer Nick The nick is removed when DNA ligase joins (ligates) the DNA fragments. © 1999 Timothy G. Standish
Mutation When Mistakes Are Made 5’ 3’ DNA Pol. 5’ ism M 5’ h c at 3’ to 5’ Exonuclease activity 5’ DNA Pol. 3’ 5’ © 1999 Timothy G. Standish
Mutation Excision Repair 3’ 3’ 5’ 5’ 3’ 3’ Th Di imi m ne er 5’ Endo. Nuclease 5’ Nicks 5’ DNA Ligase Pol. 3’ DNA Pol. 3’ 5’ Liga se © 1999 Timothy G. Standish
© 1999 Timothy G. Standish
� Question: Problem 1 – If an organisms DNA is 32 % adenine, what percent guanine, thymine, and cytosine are found in the DNA? � Answer: – As adenine always pairs with thymine, there must be 32 % thymine – % GC = 100 % - (T% + A%) = 100 % - (32 % + 32 %) = 36 % – The proportion of guanine to cytosine has to be equal as they pair with one another thus G and C % = 36 % / 2 = 18 % – G = 18 %, T = 32 % and C = 18 % © 1999 Timothy G. Standish
Problem 2 � Question: – Given the following sequence of one strand of DNA, write out the complimentary strand. – 5’AATACGCGATGCTGGTATC 3’ � Answer: – 5’AATACGCGATGCTGGTATC 3’ – 3’TTATGCGCTACGACCATAG 5’ © 1999 Timothy G. Standish
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