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Structure of DNA

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Slide 1 : APT ACADEMIC SOLUTIONS believes in people who will deliver the right solutions, and it proudly possesses them.

Dr. Pankaj KumarDirector – Apt Academic Solutions : Dr. Pankaj KumarDirector – Apt Academic Solutions Your Mentor

Slide 3 :

Slide 4 : Nature of Gene : Its expression & Regulation

Session objectives : Session objectives

Slide 6 : Evidences for DNA as genetic material

Slide 7 : Griffith’s and Avery’s Transformation Experiment Griffith and Avery provided the first direct evidence that the DNA transmits genetic information. Griffith (1928) experimented with Diplococcus pneumoniae Two different strains – S-type: Smooth, virulent strain, protected by a polysaccharide capsule and causes pneumonia R-type: Rough, harmless strain without a capsule

Slide 8 : Griffith’s Experiment (1928)

Griffith’s Experiment : Griffith’s Experiment

Griffith’s Experiment : Griffith’s Experiment Griffith’s conclusion: Some genetic material from the dead, smooth bacteria had entered the living rough bacteria and transformed the latter into smooth bacteria. The event is termed as Transformation He could not explain the exact reason for transformation

Avery Experiment : Avery Experiment

Avery Experiment : Avery Experiment O. T. Avery, C. Macleod and M. McCarty (1944) separated the extract of smooth, virulent bacteria into protein, DNA and carbohydrate fractions Each fraction was separately added to a culture medium containing live rough bacteria Only the culture that received the DNA fraction of the extract from virulent bacteria produced smooth bacteria This proved that DNA was the transforming agent When DNA fraction was treated with deoxyribonuclease, it became inactive and incapable of transforming the rough strain into the smooth strain This confirmed that DNA is the genetic material

DNA – Chemical Composition : DNA – Chemical Composition Sugar Nitrogenous Base Phosphate

Sugar : Sugar Deoxyribose sugar Ribose sugar 5-C structure O2 – bridge between carbon no. 1 & 4 OH at carbon no. 2 is replaced by H Occur in DNA 5-C structure O2 – bridge between carbon no. 1 & 4 Occur in RNA

Slide 15 : Nitrogenous Base Purine Pyrimidine 9-membered double ring structure N is present at carbon no. 1,3,7 & 9 6-membered single ring structure N is present at carbon no. 1 & 3

Purine : Purine Adenine Guanine Amino group is present at carbon no. 6 Amino group is present at carbon no. 2 Keto group is present at carbon no. 6

Pyrimidine : Pyrimidine Amino group is present at carbon no. 6 in Cytocine Methyl gropup is present at carbon no. 5 in Thymine

Phosphate : Phosphate The phosphate unit can be represented either as a phosphate ion or as phosphoric acid molecule

DNA Structure : DNA Structure Step 1: Nucleoside Sugar + Nitrogenous base = Nucleoside Nitrogenous bases are added at carbon no. 1 of sugar

DNA Structure : DNA Structure Step 2: Nucleotide Nucleoside + P = Nucleotide Phosphate is added at carbon no. 5 of sugar Adenosine tri phosphate (ATP)

DNA Structure : DNA Structure Step 3: Dinucleotide + Phosphodiester bond is formed between carbon no 3 & carbon no. 5 of nucleotide

DNA Structure : DNA Structure Step 4: Polynucleotide Polynucleotide chain PDE Bond

DNA Structure : DNA Structure Step 5: Two Polynucleotide chain joined by H-bond

DNA strand is Antiparallel : DNA strand is Antiparallel One polynucleotide chain grow in 5’ to 3’ direction Other chain is in 3’ to 5’ Antiparallel nature facilitate H-bonding between the nitrogenous bases of the two polynucleotide chain

Slide 25 : DNA Structure Why Purine pairs with Pyrimidine ? Distance between two DNA strand is fixed (20 A0) Gap of 20 A0 can accommodate only three ring structure It can happen only if purine pairs with pyrimidne

DNA Structure : DNA Structure Adenine pairs with Thymine by two hydrogen bond Guanine pairs with Cytocine by three hydrogen bond Adenine pairs with Uracil AT & GC pairing occurs to satisfy the bonding configuration between nucleotides

Watson & Crick Model : Watson & Crick Model DNA double helix model is the most accepted model Model is based on X-ray crystallographic study of Wilkins & Franklin Chemical analysis of Chargaff

X-ray crystallographic study of Wilkins & Franklin show that : X-ray crystallographic study of Wilkins & Franklin show that Illustrative Problem DNA is helical structure DNA is of uniform diameter DNA is made up of purine & pyrimidine All of the above

Chargaff’s Rule : Chargaff’s Rule A = T G = C Purine = Pyrimidine Hence, A + G = T + C k = specific for every species

B-DNA : B-DNA Most common form of DNA Sugar phosphate forms a backbone Distance between two strand or diameter is 20 A0 Vertical height or one complete turn is 34 A0 (Distance between two consecutive major / minor groove) No. of base pair in one turn = 10 Distance between two consecutive base pair is 3.4 A0 Angular displacement between two consecutive base = 360

B-DNA : B-DNA What is the distance between major & minor groove of B-DNA ? 17 A0

Forms of DNA : Forms of DNA Homework

Forms of DNA : Forms of DNA

Slide 34 : Do U find any difference between major & minor groove of A & B DNA ? Grooves are well differentiated in B-DNA & less differentiated in A-DNA

Slide 35 : Grooves are not differentiated

Other forms of DNA : Other forms of DNA Palindromic DNA c DNA Selfish DNA Housekeeping DNA Happy net surfing !!

DNA : DNA RNA

Home Work : Home Work Give two examples in which single stranded DNA & double stranded RNA occurs

Slide 39 : Structure of RNA

t-RNA : t-RNA Clover leaf model It is 2-D structure Presence of three loops Anticodon loop has 7 base pairs, three of them act as anticodon G is present at 5’ end ACC is present at unpaired 3’ end Amino acid binds at 3’ end

t-RNA : t-RNA 3-D structure is given by S.H.Kim It is L-shaped

Hn-RNA (Heterogenous RNA) : Hn-RNA (Heterogenous RNA)

Slide 43 :

Slide 44 : Hn-RNA is the immediate transcriptional product in eukaryote It consists of alternate sequence of intron & exon Introns are non-coding & exons are coding sequences During post-transcriptonal modifications introns are removed m-RNA consists of exons only

Home work : Home work Why post transcriptional modifications are not required in prokaryote ? Rush your Answer to pk@aptacads.com

Sn-RNA (small nuclear RNA) : Sn-RNA (small nuclear RNA) Immediate transcriptional product in eukaryote No post-transcriptional modifications Helps in the formation of spliceosome Sc-RNA (small cytoplasmic RNA) Immediate transcriptional product in eukaryote No post-transcriptional modifications Helps in the formation of signal recognition protein (SRP) during protein synthesis

Slide 47 : Replication of DNA Self duplication process of DNA Semi-conservative in nature Replication produce molecules with both old and new DNA Each molecule consists of one old strand and one new one, hence called as semi-conservative

Slide 48 : Experimental evidence for semi-conservative replication: Meselson and Stahl (1958) provided experimental evidence supporting semi-conservative mode of DNA replication in E. coli by DNA banding techinique Taylor provided experimental evidence supporting semi-conservative mode of DNA replication in Vicia faba by autoradiographic technique Semi-conservative replication of DNA

Slide 49 : Meselson and Stahl’s experiment

E. coli about to replicate was placed in a medium containing radioactive thymidine for five minutes. Then it was made to replicate in a normal medium. Which one of the following observation shall be correct? : E. coli about to replicate was placed in a medium containing radioactive thymidine for five minutes. Then it was made to replicate in a normal medium. Which one of the following observation shall be correct? Both the strands of DNA will be radioactive One strand radioactive Each strand half radioactive None is radioactive Illustrative Problem

E. coli was allowed to grow in a medium containing radioactive thymidine for one month. Then it was made to replicate in a normal medium for 40 minutes. Number of non radioactive DNA strand is : E. coli was allowed to grow in a medium containing radioactive thymidine for one month. Then it was made to replicate in a normal medium for 40 minutes. Number of non radioactive DNA strand is 2 4 6 8 Illustrative Problem Red – radioactve Green –non-radoactive E.coli replicates every 20 minutes No. of non-radioactive (green) strand is 6

DNA Replication : DNA Replication Replication occurs in replication fork or replication bubble Prokaryotes are mono repliconic Eukaryotes are multi-repliconic

Slide 53 :

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