Download - Genetics Finals

1

Linkage and Chromosome Mapping

in Eukaryotes

Fundamental GeneticsLecture 8

John Donnie A. Ramos, Ph.D. Dept. of Biological Sciences

College of Science University of Santo Tomas

Linked Genes

Genes located in the same chromosomes

Initiated by Thomas Morgan and Alfred Sturtevant

Will not segregate independently

Affected by crossing over

2

Linkage vs Independent Assortment

Linked Genes in DrosophilaRed eyes (bw+) dominant to mutant brown eyes (bw)

Thin wing veins (hv+) dominant to mutant heavy wing veins (hv)

3

X-Linked Genes in DrosophilaCross A

Gray body (y+) dominant to mutant yellow body (y)

Red eyes (w+) dominant to mutant white eyes (w)

Cross B

Red eyes (w+) dominant to mutant white eyes (w)

Normal wings (m+) dominant to mutant miniature wings (m)

Due to linkage and crossing over that occurred during meiosis

Distance between linked genes is related degree of crossing over

Chromosome MappingDetermining the distances between genes and the order of sequence in a chromosomeUses the frequency of crossing

The shorter the distance between linked genes, the lower the frequency of crossing-over.The longer the distance, the higher the frequency of crossing over to occur.

Frequencies of crossing over:1. Yellow, white 00.5 %2. White, Miniature 34.0 %3. Yellow, miniature 35.4 %

1% of crossing over = 1 map unit (centimorgan, cM)

4

Distance Affects Crossover

Single Crossover

50% are recombinant gametes50 % non-crossover gametesIf distance between genes is more than 50 map units, ~100 % crossing over will occur.

5

Multiple Crossover

Occurrence of more than one crossovers between non-sister chromatids.Produces double crossover (DCO) gametesIf the probability of crossover between A and B is 20% (0.20) and the probility of crossover between B and C is 30% (0.30), the frequency of DCO is 6 % (0.06) Product Law: (0.2)(0.3)=0.06

Three-Point Mapping

The percentage of crossing over could be used to map genes in a chromosomeThree criteria needed for successful mapping:

Genotypes of organisms producing the crossover gametes must be heterozygous for all gene loci Cross must be constructed so that the genotypes of gametes could be determined based on the phenotypes of the offspring.Large number of offspring must be produced

6

Traits considered:

1. Body color

Gray(y+) dominant to yellow (y)

2. Eye color

Red eyes (w+) dominant to white (w)

3. Eye shape

Normal (ec+) dominant to echinus (ec)

10,000

Determining Gene SequenceSteps:

Determine 3 possible ordersw-y-ec (y at the middle)y-ec- w (ec at the middle)y-w-ec (w at the middle)

Perform a theoretical double cross overCompare the theoretical DCO with actual DCO (least no.)Perform theoretical NCOI and NCOII and compare with data

White echinus eyes, gray body

Red normal eyes, yellow body

Yellow body, normal white eyes

Gray body, echinus red eyes

Yellow body, echinus red eyes

Gray body, white normal eyes

7

Unknown Gene Sequence

Total=1109

Unknown Gene Sequence

8

Not all crossovers can be detected

Degree of inaccuracy increases with increasing distance between linked genes

Observed vs Expected DCO

Observed DCO = double cross-over that actually occurred

Example: (44 + 42)/1109 = 0.078

Expected DCO = theoretical double crossoversProduct of all the SCOI and SCOIIExample: (82+79+44+42) / 1109 = 0.223

(200+195+44+42) / 1109 = 0.434DCO exp= (0.223)(0.434) = 0.097

9

Coefficient of Coincidence and Interference

Coefficient of Coincidence (C)The measure of actual DCOs that occurredC = Observed DCO / Expected DCO

= 0.078/0.097= 0.804 or 80.4%

Interference (I)phenomenon when a crossover event in one region of a

chromosome inhibits a second event to occur in a nearby region)

I = 1-C= 1-0.804 = 0.196 or 19.6%

Interpretation: 19.6% of expected DCO did not occur or only 80.4% of expected DCO was observed

Problem 1

A stock of corn homozygous for the recessive linked genes colorless (c), shrunken (sh), and waxy (wx) was crossed to a stock of homozygous for the dominant wild type alleles of the genes (+ + +). The F1 plants were then backcrossed to the homozygous recessive stock. The F2 results were as follows:

Phenotype Number Phenotype Number

+ + + 17,959 + + wx 4,455

c sh wx 17,699 c sh + 4,654

+ sh wx 509 + sh + 20

c + + 524 c + wx 12

a. Determine the distance between the c and sh

b. Determine the distance between the sh and wx

c. Determine the distance between c and wx

d. Give the coefficient of coincidence

e. Compute for the interference

10

Problem 1: Solutionc sh wxc sh wx

+ + ++ + +x

x c sh wxc sh wx

c sh wx+ + +

NCO + + + = 17,959c sh wx = 17,699

SCOI + sh wx = 509c + + = 524

SCOII + + wx = 4,455c sh + = 4,654

DCO + sh + = 20c + wx = 12Total = 45,832

35,658

1,033

9,109

32

=

=

=

= 00.07 %

19.87 %

02.25 %

77.80 %

Problem 1: Solution

Distance between c and sh = (509 + 524 + 20 +12) / 45,832

= 0.0232 or 2.32 %

Distance between sh and wx = (4466 + 4654 + 20 + 12) / 45832

= 0.1994 or 19.94 %

Distance between c and wx = 2.32 + 19.94

= 22.26

11

Problem 1: Solution

c sh wx

2.32 mu 19.94 mu

22.26 mu

C = (0.0007) / (0.0232)(0.1984) = 0.1521 or 15.21%

I = 1-C= 1-0.1521= 0.8479 or 84.79 %

Problem 2

In a variety of tomato plant, the mutant genes o (oblate fruit), h (hairy fruit), and c (compound inflorescence) are all located in chromosome 2. These genes are recessive to their wild type alleles round fruit, hairless and single inflorescence, respectively. A testcross mating of an F1 heterozygote for all three genes resulted in the following phenotypes:

Phenotypes Number Phenotypes Number

+ + + 73 + + c 348

+ h + 2 + h c 96

o + + 110 o + c 2

o h + 306 o h c 63

a. Determine the sequence of the 3 genes in chromosome 2

b. Give genotypes of the homozygous parents (P1) used in making the F1 heterozygote.

c. Compute for the map distances between the genes

d. Give the coefficient of coincidence and interference

12

Problem 2: Solution+ + +

o h c

o h c

o h cx

NCO: o h + = 306

+ + c = 348

DCO: + h + = 2

o + c = 2

Inference from given data:

Sequence of genes is not correct

One chromosome contains 2 wild type alleles while the homolog contains the 3rd wild type allele

Three possible orders of the genes involved:

o h c

o c h

h o c

Find a sequence that will satisfy both NCO and DCO

o h ++ + c Satisfies NCO but not DCO

+ c ho + + Satisfies DCO but not NCO

+ o hc + +

Satisfies both NCO and DCO

Problem 2: Solution

+ o h

c + +

o h c

o h cx

NCO: o h + (same as + o h) = 306+ + c (same as c + +) = 348

DCO: + h + (same as + + h) = 2o + c (same as c o +) = 2

Try if the sequence can satisfy the SCOs

SCO I: + + + = 73o h c (same as c o h) = 63

SCOII: o + + (same as + o +) = 110+ h c (same as c + h) = 96

Total = 1,000

654 = 0.654 or 65.4%

4 = 0.004 or 0.4%

136 = 0.136 or 13.6%

206 = 0.206 or 20.6%

13

Problem 1: Solution

Distance between c and o = (73 + 63 + 2 + 2) / 1,000

= 0.140 or 14 % / cM

Distance between o and h = (110 + 96 + 20 + 12) / 1000

= 0.210 or 21 % / cM

Distance between c and wx = 14 + 21

= 35 cM

Problem 1: Solution

c o h

14 cM 21.0 cM

35 cM

C = (0.004) / (0.14)(0.21) = 0.1361 or 13.61%

I = 1-C= 1-0.1361= 0.8639 or 86.39 %

1

DNA Replication and Synthesis




The Flow of Biological Information

DNA

RNA

Protein

Replication

Transcription

Translation

2

Modes of DNA Replication

Semiconservative Replication

3

Semiconservative Replication in Prokaryotes

Mathew Messelson and Franklin Stahl (1958)15N – heavy isotope of N (contains 1 more neutron) compared to 14N15N has high sedimentation rate in cesium chloride compared to 14N

Semiconservative Replication in Prokaryotes

Expected results of the Messelson-Stahl experiment

4

Semiconservative Replication in Eukaryotes

J. Herbert Taylor, Philip Woods, and Walter Hughes (1957)

Used root tip cells from Viciafaba (broad bean)

Monitored replication using 3H-Thymidine to label DNA

Used autoradiography to determine the incorporation of 3H-Thymidine

Arrested cells at metaphase using colchicine

Replication of E. coli Plasmid

Shown by John Cairns (1981) using radioisotopes and radiography

Replication starts in a single OriC –origin of replication (245 bp)

Replication is bidirectional

Replication fork – unwound DNA helix

Replicon – replicated DNA

Ter region – region of replication termination

5

DNA Synthesis in MicroorganismsDNA polymerase I (928 aa) – catalyses the synthesis of DNA in vitro(A. Kornberg, 1957)

Requirements:

Deoxyribonucleoside triphosphates, dNTPs (dATP, dCTP, dGTP, dTTP)

DNA template

Primer

Chain Elongation5’ to 3’ direction of DNA synthesis (requires 3’ end of the DNA template)

Each step incorporates free 3’ OH group for further elongation

DNA replication using DNA polymerase is of high fidelity (highlyaccurate)

With exonuclease activity (proofreading ability)

6

DNA PolymerasesAll 3 types requires a primer

Complex proteins (100,000 Da)

Functions of DNA polymerases in vivo

DNA Pol I – proofreading; removes primers and fills gaps

DNA Pol II - mainly involved in DNA repair from external damage

DNA Pol III – main enzyme involved in DNA synthesis

a holoenzyme (>600,000 Da) – forms replisome when attached to a replication fork.

Replication in Prokaryotes1. Unwinding of DNA helix

2. Initiation of DNA synthesis

3. DNA synthesis proper (elongation)

4. Sealing gaps

5. Proofreading and error correction

7

Unwinding of DNA Helix

Takes place in oriC (245 bp) –repeating 9mers and 13mers

Function of helicases (Dna A, B, C) – requires ATP hydrolysis to break hydrogen bonds

Initiated by Dna A – binds to 9mers

Binding of Dna B and Dna C to unwound helix

Single-stranded binding proteins(SSBPs) – prevents reannealing of replication bubble.

DNA gyrase (a DNA topoisomerase) – relaxes the supercoiling of DNA helix

Initiation of DNA Synthesis

Synthesis of RNA primer – 5 to 15 RNA bases complementary to the DNA template

Catalysed by primase (an RNA polymerase)

Pimase does not require free 3’ end to initiate synthesis (not unlike DNA polymerase III)

Function of primase will be continued by DNA polymerase III.

8

DNA Synthesis (Elongation)Function of DNA polymerase III

Requires free 3’ end

Direction of elongation: 5’ to 3’

DNA synthesis is continuous in 3’ to 5’DNA strand (leading strand) and discontinuous in the 5’ to 3’ DNA strand (lagging strand).

Okazaki fragments – short DNA fragments produced in the lagging strand

Concurrent synthesis of leading and lagging strands occur by using DNA poldimer and by a looping mechanism for the lagging strand

Sealing of Gaps, Proofreading and Error Correction

DNA polymerase I removes all RNA bases produced by primase (creates gaps in the lagging strand) and replaces it with DNA bases (U to T).

DNA ligase seals the gaps by forming phosphodiester bonds

Exonuclease proofreading (identification of mismatched bases) is a function of both DNA polemerase I and III (both with 3’-5’ exonucleaseactivity)

ε subunit of DNA polymerase III is involved in proofreading.

Assures high fidelity of DNA replication

9

Mutations Affect Replication

Replication in Eukaryotes

Presence of multiple replication origin (faster replication, guarantees replication of a big genome) – 25K replicons in mammalian cells

Autonomously replicating sequences (ARSs) – origin of replication in yeasts (11 bp)

Origin site is AT rich region

Helicase unwinds double stranded DNA and removes histone proteins from DNA

Histones reassociates while DNA synthesis occurs.

10

Eukaryotic DNA PolymerasesPol α - initiates nuclear DNA synthesis

4 subunits (2 acts primase – produces RNA primers)Acts on both leading and lagging strands2 other subunits continue elongation step (DNA synthesis)Low processivity (short length of synthesized DNA prior to dissociation)

Pol δ - replaces Pol α (called polymerase switching)High processivity (during elongation)With 3’-5’ exonuclease activity (proofreading)

Pol ε - nuclear DNA synthesisPol β - DNA repair (the only eukaryotic DNA polymerase with single subunit)Pol ξ - DNA repairPol γ - mitochondrial DNA synthesis (encoded by nuclear gene)

Eukaryotes has a high copy number of DNA polymerases (ex. Pol αmay be up to 50K copies)

Eukaryotic DNA Replication

Telomeres – linear ends of eukaryotic chromosomes

Problem with lagging strand: no 3’ needed by DNA polymerase I (after removal of RNA primers)

Possible result: chromosome with shorter lagging strand every replication step

11

TelomeraseEnzyme that adds TTGGGG repeats on the telomeres (first identified in Tetrahymena)

Prevents shortening of chromosomes

Forms a “hairpin loop” on chromosome ends using G-G bonds

Creates a free 3’ on lagging strand that can be used by DNA polymerase I to replaced the removed RNA primer

Telomerase is a ribonucleoprotein and contains RNA sequence (5’ AACCCC 3”-serving as template) – reverse transcriptase

Cleavage of loop after DNA synthesis

DNA Recombination

Exchange of genetic material

Homologous recombination

Ex. Rec A protein (produces recB, recCand recD genes)

12

Gene Conversion

Exchange of genetic information between non-homologous chromosomes

Type of chromosome mutation (recombination)

First identified in Neurospora (by Mary Mitchell)

Can be repaired but forms recombined genetic material

1

The Genetic Code and Transcription




Flow of Genetic Information

2

Linear form (mRNA derived from DNA)

Triplet codons (triplets of ribonucleotides coding for 1 amino acid)

Unambiguous (1 codon = 1 amino acid only)

Degenerate ( 1 amino acid can be specified by several codons)

Contains specific start and stop codons

Commaless (no breaks once translation starts until the stop codon is reached)

Non-overlapping (single reading frame)

Universal (same ribonucleotide used by all organisms)

The Genetic Code

Francois Jacob and Jacques Monod(1961) – messenger RNA (mRNA)

Sydney Brenner (1960s) – codon in triplets (minimal use of the 4 mRNA bases to specificy 20 aa) (43=64)

Francis Crick – frameshift mutations alters the codons

Mariane Manago and Severo Ochoa -polynucleotide phosphorylase(synthesis of RNA without template)-paved the way to the production of RNA polymeres in cell free-systems

The Discovery of the Genetic Code

3

Marshall Nirenberg and J. Heinrich Matthaei (1661) – codonsused cell-free protein synthesizing system and polynucleotide phosphorylaseRNA Homopolymers (UUUUUU…, AAAAAAA…, CCCCCC…, GGGGG…)

UUU (Phenylalanine)AAA (Lysine)CCC (Proline)

RNA Mixed Copolymers

1A:5C (1/6 A: 5/6C)

The Discovery of the Genetic Code

Developed by M. Nirenberg and P. Leder (1964)Mimics the in vivo translation of proteins where a mRNA-tRNA-ribosome complex is formed when all three macromolecules are allowed to interact.

The Triplet Binding Assay

4

Developed by Gobind Khorana (1960s)Synthetic long RNAs with repeating sequences

Repeating Copolymers

Results of Repeating Copolymers

5

The Universal Genetic Code

Degeneracy

Wobble HypothesisStart codon

(N-formylmethionine)

Termination codonsUniversalVirusesBacteriaArchaeaEukaryotes

Exceptions to the Universal Code

6

TranscriptionUses DNA as a template

Catalyzed by RNA polymerase(holoenzyme of 500 kD)

αββ’σ subunits

Sense strand / template strand –DNA strand used as a template for transcription

Promoter region – DNA sequence recognized by σ factor to initiate transcription (60 bases). (upstream of a gene)

TATA box (Pribnow box) –TATAAT sequence

Sigma factor (σ70, σ28, σ32, σ54)

TranscriptionRNA polymerase don’t need primers

Elongation in 5’ to 3’ direction

Rate in E coli: 50 bases/sec at 37°C

Termination is a function of rho(ρ) factor – hexameric protein interacting with the end of a gene

Polycistronic mRNA – bacterial mRNA containing information for the synthesis of proteins of related function

Monocystronic mRNA –eukaryotic mRNA containing information for a single protein.

7

Eukaryotic TranscriptionFeatures of eukaryotic transcription different from prokaryotic transcription:

Transcription inside the nucleus under the direction of 3 different RNA polymerases

Presence of protein factors (promoters, enhancers, etc.) binding to the upstream portion of a gene (cis-acting elements) during initiation step.

Presence of post-transcriptional regulation.

Cis -acting Elements

TATA Box (Goldberg-Hogness Box)

Located 30 bases upstream from the start of transcription (-30)

Consensus sequence: TATAAAA

Facilitates denaturation of helix because it is AT-rich region

CAAT Box

Located 80 bases upstream from the start of transcription (-80)

Consensus sequence: GGCCAATCT

Influence the efficiency of the promoter

8

Trans -acting Factors

Transcription factors – facilitates template binding during the initiation of transcription

Example:

TFIID (TATA-binding protein or TBP) – binds to TATA-box

Post-transcriptional Processing

7-methylguanosine cap (7mG)Protection from nucleasesRole mRNA transport across the nuclear membrane

3’ cleavage site:AAUAAAFailure of 3’ cleavage results to absence of poly A tail

Split genes – contains intervening sequences

9

RNA Splicing

Ribozyme – RNA with catalytic activity

Self-excision process –process of RNA splicing or intron removal.

Transesterification –interaction between guanosine and the transcript.

2 successive transesterification processes

The Spliceosome

Alternative splicing

Small nuclear ribonucleoproteins(snRNP or snurps) – bonds to GU or AG sites of introns

2 transesterification processes

Snurps form a loop (lariat) in the branch point region

Produces isoforms of proteins

10

RNA Editing

Substitution editing

changes in the nucleotide bases of a given mRNA

Common in mitochondrial RNA and chloroplast RNA

Example: Apoliprotein B (Apo B) – C to U change CAA to UAA

Insertion / deletion editing

addition or removal of nucleotide sequences

Common in mitochondrial RNA or guide RNA (gRNA)

1

Translation and Proteins


John Donnie A. Ramos, Ph.D.Department of Biological Sciences


The Products of Transcription

Messenger RNA (mRNA)primary structurelinear sequence of RNA basescarries the genetic information in

the form of codons

Ribosomal RNA (rRNA)assumes a 3D structure

(complexed with proteins)site of protein synthesis

Transfer RNA (tRNA)assumes a cloverleaf structurecarries amino acids from

cytoplasm to ribosomes

Codons

2

RibosomeEncoded by rDNA(ribosomal gene)

Synthesized by RNA polymerase I in the nucleolus

Complex of RNA and proteins (monosome)

Prokaryotes: 10K/cell

Transfer RNA75-90 nucleotides

Nucleotides are post-transcriptionallymodified

2D cloverleaf structure (Rodbert Holley) due to base pairing

Amino acid binding site - ends in CCA3’ and 5’GAnticodon loop – contains RNA bases complementary to the codonsOther loops serves as recognition sites for enzymes during translation

3D Structure

2D Structure

3

Steps in Translation1.Charging of tRNA

2.Initiation of translation

3.Elongation of polypeptide chain

4.Termination of translation

Loading of specific amino acid to its own tRNACatalyzed by aminoacyl tRNA synthetase32 different tRNA (despite the presence of 61 codons (bec. Of wobbling mechanisms)20 different aminoacyl tRNA synthetasesIsoaccepting tRNA – tRNA that binds to aaEnd product: aminoacyl-tRNA complex

Charging of tRNA

Initiation of Translation

Shine-Dalgarmo sequence (5’AGGAGG3’) – sequence that precedes the first codonin prokaryote m RNAFormylmethionine (fmet) – the first amino acid of most polypeptides

4

Elongation of Polypeptide Chain

Peptidyl site (P site) – contains the elongating peptideAminoacyl site (A site) – contains the amino acid to be addedExit site (E site) – exit of uncharged tRNAPeptidyl transferase – catalyzes the formation of peptide bondHigh efficiency (error rate 10-4)Rate of elongation: 15 aa/sec at 37°C (E. coli)

Protein Factors Involved in Translation

5

Termination of Translation

Signaled by stop codons (UAG, UAA, UGA)Release Factor 1 (RF1) – recognizes stop codon UAA and UAGRelease Factor 2 (RF2) – recognizes stop codons UGA and UAARelease factors are GTP dependentPost-translational modification starts after release from ribosome

PolyribosomesSingle mRNA being used by different ribosomes for the process of translationAlso called polysomesA mechanism to produce more polypeptide (protein) copies

Translation of hemoglobin mRNA in rabbit reticulocyte

Translation in giant salivary gland cells of midgefly

6

Translation in EukaryotesmRNAs stays in the cytoplasm for longer periods before degradation by RNAses (hours)Ribosomes are much biggermRNA is capped with 7-methyguanosine (7MG)mRNA contains an initiation sequence called Kozak sequence (ACCAUGG) discovered by Marilyn KozakFormylmethionine (fMet) is not required for initiation but met is often used as a start codonMore complex protein factors involved in different stepsElongating polypeptide enters the ER immediately as translation occurs

Proteins Form PhenotypesPhenyketonuria

Mental retardationAutosomalrecessiveInability of Phe to converted to TyrAccumulation of Phe and its derivatives in cerobrospinal fluid

AlkaptonuriaAutosomal recessiveDarkening ears and noseBenign arthritic conditions

7

Genes and ProteinsOne gene: one enzyme hypothesis

Proposed by George Beadle and Edward Tatum (1940s)Experiments in Neurosporamutants

Genes and Proteins

One gene: one protein (polypeptide chain)Not all protein are enzymes Example: Sickle Cell Anemia (mutant hemoglobin)

8

Amino Acids

Protein Structure

Primary Structure Secondary Structure

Tertiary StructureQuaternary Structure

9

Post-translational Modifications

Cleavage of formylmethionineCleavage of signal peptidesAcetylation of amino groupPhosphorylation of certain amino acidsGlycosylationTrimmining of polypetidesAddition of metallic groups

Molecular Chaperons – help proteins undergo correct protein folding to become functional molecules.

Protein Function

Structural FunctionCollagenKeratinActinMyosin

Regulatory FunctionHormonesHemoglobinMyoglobin

Defense FunctionAntibodiesComplement proteins

Catalytic FunctionEnzymesRibozymes

OthersHistonesReceptors

Enzyme Activity

10

Protein Domains and Exon Shuffling

DNA organization of an LDL receptor gene

Structural domains of a fibronnectin molecule

1

DNA Structure and Analysis




DNA: The String of Life

James Watson Francis Crick

2

Characteristics of the Genetic Material

Replication

Storage of information

Expression of information

Variation by mutation

Central Dogma of Molecular Genetics

Early Studies on the Genetic Material

Friedrick Miescher (1868) – acid substance from nuclei called nuclein

Phoebus Levene (1910) – tetranucleotide hypothesis (equal amount of nucleotides)

Frederick Griffith (1927) – In vivo transformation experiment

Oswald Avery, Colin MacLeod, Maclyn McCarty (1944) – In vitro transformation experiment (bacteriophage)

Alfred Hershey, Martha Chase (1952) – Bacteriophagetransformation

William Astbury (1938) – X-ray diffraction analysis of DNA

Rosalind Franklin (1950) – improved X-ray diffraction analysis of DNA

James Watson and Francis Crick (1953) – DNA double helix structure

3

In Vivo Transformation Experiment

“Transformation might be due to the polysaccharide capsule or some compound required for capsule synthesis”

In Vitro Transformation Experiment

DNA is responsible for the transformation of avirulent strain to a virulent type!

4

Hershey-Chase Experiment

DNA (and not protein) is the genetic material in phage T2.

Evidences Favoring DNA as the Genetic Material

DNA is found only where genetic function is known to occur but protein is ubiquitous.

DNA content of cells is directly correlated with the number of sets of chromosomes present but not for proteins

5

Evidences Favoring DNA as the Genetic Material

DNA absorbs UV at the same wavelength where mutation occurs (action spectrum) but proteins absorbs at different wavelength

Recombinant DNA Technology (transgenic organisms) – direct evidence

RNA: Genetic Material in Some Viruses

First identified in 1956 in tobacco mosaic virus (TMV)

Uses RNA replicase to duplicate genetic material

Retroviruses – undergo reverse transcription (RNA to cDNA) using reverse transcriptase

6

DNA Structure

Proposed by Watson and Crick in 1953 based on:

Base composition analysis of hydrolyzed samples of DNA

X-ray diffraction studies of DNA

Sequence of nucleotides codes for the genetic information (4n where n refers to the no. of nucleotides)

DNA Structure

7

DNA Structure

Precursor molecule in nucleic acid synthesis

Source of energy (ATP)

Nucleotide Linkage

8

Base Composition StudiesFirst studied by Erwin Chargaff (1949-1953)

Agrees with Watson and Crick DNA model

Chargaff Rule

Amount of A is proportional to T while C is proportional to G

Sum of purines (A+G) equal to sum of pyrimidines (C + T)

Percentage of G + C does not necessarily equal to percentage of A + T

9

The Watson-Crick DNA Model

Right-handed double helix

Antiparallel chains

Nitrogenous bases as flat structures inside the helix

Bases are 3.4 A apart

Base complementarity (A-T and G-C)

10 bases every 360° turn

34 A every complete turn

Double helix diameter is 20 A

Semiconservative mode of replication

Types of DNA

AbsentModifiedPresentMajor groove

Left-handedRight-handedRight-handedDirection of turn

18 A23 A20 ADiameter of helix

40.837.4 A34 ALength / 360° turn

12 bp11 bp10 bpBases / 360° turn

Z DNAA DNAB DNACriteria

10

RNA StructureRibose sugarSame nitrogenous bases as DNA except that T replaced by USingle stranded (but can form double strands)Forms:

Ribosomal RNA (rRNA)Messenger RNA (mRNA)Transfer RNA (tRNA)

Differs by sedimentation rate (Svedverg Coefficient)

Small Nuclear RNA (snRNA)Telomerase RNAAntisense RNA

Nucleic Acid Unique CharacteristicsHydrogen bonds breaks at high temperature (denaturation or unwinding)Hydrogen bonds reform at lower temperature (annealing)Melting Temperature (Tm)= temperature at which 50 % of H bonds are broken (DNA with higher GC content has higher Tm)Can be measured using spectrophotometer (absorbance at 260 nm)With increasing temperature, the viscosity of DNA decreases and UV absorption increase

11

Molecular Hybridization

Annealing of nucleic acid (DNA or RNA) strands sharing nucleotide sequence similarity

Used to identify homologous genes in different species

Example: In situ hybridization or Fluorescence in situ hybridization (FISH)

Reassociation KineticsMeasures the rate of annealing between complementary strands

Measures half reaction time (point when ½ of the reaction are double stranded)

Half Reaction is lower in smaller genomes

Used to measure repetitive DNAsequences (characteristic of eukaryotes)

12

Electrophoresis

Agarose gel electrophoresisPolyacrylaminde gel electrophoresisSeparates nucleic acids by size under an electrical fieldDNA is negatively charged (travels to + charge)Southern Blot –detection of DNANorthern Blot –detection of RNA