BNG5113 - Lecture 1

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Introduction

BNG5113

Molecular and Cellular Bioengineering

Spring 2021

Assist. Prof. Elif Eren

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Are there any topics you want to study

during this molecular and cellular

bioengineering course?

https://www.menti.com/bcgyqw8sjd The voting code is 4688 5373

bionanotechnology protein covid protein system genomics drug delivery virology engineering in vitro fertilization cancer nanotechnology cellular chemical balance genetic engineering signal transduction genes Q) c gene expression biotechnology gene therapy

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Aim of the Course

To provide the molecular and cellular bases of life

from an engineering perspective.

Molecular

Biology

and Genetics

Bioengineering Biomedical Engineering

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Week/Place Course Topic To Do Assignments & Deadline W1 March 8, 2021 ONLINE DNA replication • Read the lecture notes of W1 W2 March 15, 2021 ONLINE DNA elasticity • Study lectures of W1 • Read the lecture notes of W2 W3 March 22, 2021 ONLINE DNA sequencing technologies I • Study lectures of W1-2 • Read the lecture notes of W3 W4 March 29, 2021 ONLINE DNA sequencing technologies II • Study lectures of W1-3 • Read the lecture notes of W4 W5 April 5, 2021 ONLINE Melting and other structural transitions in DNA • Study lectures of W1-4 • Read the lecture notes of W5 W6 April 12, 2021 ONLINE DNA nanotechnology • Study lectures of W1-5 • Read the lecture notes of W6 W7 April 19, 2021 ONLINE Driving forces in protein folding and binding I • Study lectures of W1-6 • Read the lecture notes of W7 W8 April 26, 2021 ONLINE Driving forces in protein folding and binding II • Study lectures of W1-7 • Read the lecture notes of W8 W9 May 3, 2021 ONLINE Allostery • Study lectures of W1-8 • Read the lecture notes of W9 W10 May 10, 2021 ONLINE Protein design I • Study lecture of W1-9 • Read the lecture notes W10 W11 May 17, 2021 ONLINE Protein design II • Study lectures of W1-10 • Read the lecture notes W11 W12 May 24, 2021 ONLINE Metabolic engineering • Study lectures of W1-11 • Read the lecture notes W12 W13 May 31, 2021 ONLINE Synthetic biology I • Study lectures of W1-12 • Read the lecture notes W13 W14 June 07, 2021 ONLINE Synthetic biology II • Study lectures of W1-13 • Read the lecture notes W14 The Course

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Ø Griffiths, J. B., 1990. Animal Cell Biotechnology, Vol 1-4; Eds. R. E.

Spier, Acad. Pres Inc. Freshney I. R. 1994. Culture of Animal Cells. Wiley Liss Inc. Neumann, K., H., A.Kumar, J. Imani, 2009.

Ø Plant Cell and Tissue Culture – A Tool in Biotechnology; Basics and

Application. Springer – Verlag , Berlin Heidelberg. George, E. F., M.A. Hall, G.-J.De Klerk, 2008.

Reference Books

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Wednesday 9:30- 17:30 on appointment.

Itslearning Messaging

Office Hours

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Evaluation of the Course

Assignment Description Scoring Weight

(%)

Homework A homework will be given. Details will be provided during the course. 100 25

Midterm

The midterm exam will include all subjects covered up to date of the exam. The exact date of the midterm exam will be announced later.

100 35

Final Exam You will have a final exam covering all lectures. 100 40

TOTAL 100 100

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DNA Replication

BNG5113

Molecular and Cellular Bioengineering

Spring 2021

Assist. Prof. Elif Eren

1

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Outline

Ø PART I: Discovery of DNA

Ø PART II: DNA Structure

Ø PART III: Replication of the DNA

Ø PART IV: Applications in Bioengineering

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PART I: Discovery of DNA

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The discovery of DNA

Mendel

Miescher

Kossel Franklin

Watson/

Crick

1951 1949 1881 1869 1866

“Inheritance”

“Nuclein”

“DNA, A T C G”

“DNA

Visualization”

“Double Helix”

Chargaff

“Equal

amount of A&T, G&C”

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Gregory Mendel

https://www.sciencelearn.org.nz/resources/2000-mendel-s-principles-of-inheritance

1866

Inheritance in pea plants

Monastery of St Thomas

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Gregory Mendel

https://www.sciencelearn.org.nz/resources/2000-mendel-s-principles-of-inheritance

1866

Inheritance in pea plants

Traits:

Seed shape, Seed color, Flower color, Pod shape

Round Yellow Purple Inflated Green Wrinkled Green Wh ite Constricted Yellow

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Gregory Mendel

https://www.sciencelearn.org.nz/resources/2000-mendel-s-principles-of-inheritance

1866

Inheritance in pea plants

Traits:

Seed shape, Seed color, Flower color, Pod shape

Round Wrinkled

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Gregory Mendel

https://www.sciencelearn.org.nz/resources/2000-mendel-s-principles-of-inheritance

1866

Inheritance in pea plants

Classical genetics:

Study of genetics through the analysis

of the offspring from mating.

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Friedrich Miescher

1869

Identification of the “Nuclein”

https://www.dna-worldwide.com/resource/160/history-dna-timeline#2

White blood cells

First isolation of

nucleic acid (1869)

Precipitate Nuclein

(DNA associated with proteins)

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Albrecht Kossel

1881

DNA, nucleotides

https://www.dna-worldwide.com/resource/160/history-dna-timeline#2

Nuclein=

DNA associated with proteins

DNA and RNA are composed of 5 nucleotides:

adenine, cytosine, guanine, thymine, and uracil

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Erwin Chargaff

1949

A=T, C=G Chargaff’s Rules

1- The amount of A, T, C, G vary among species

2- The number of A is equal to the number of T;

The number of C is equal to the number of G.

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Rosalind Franklin

1950

Photographs of DNA

https://www.biography.com/scientist/rosalind-franklin

Picture 51

First Picture of DNA

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Watson/Crick and Wilkins

Maurice Wilkins

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Francis Crick James Watson Wurice Wikins

Nobel Prize in Physiology or Medicine 1962

Rosalnd Franklin

The Nobel Prize in Physiology or Medicine 1962 was awarded jointly to Francis Harry Compton Crick, James Dewey Watson and Maurice Hugh Frederick Wilkins “for their discoveries

concerning the molecular structure of nucleic acids and its significance for information

transfer in living material.”

Who discovered:

Who obtained the prize:

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The discovery of DNA

Mendel

Miescher

Kossel Franklin

Watson/

Crick

Charpentier/

Doudna

2020 1951 1949 1881 1869 1866

“Inheritance”

“Nuclein”

“DNA, A T C G”

“DNA

Visualization”

“Double Helix”

“Genome Editing”

Chargaff

“Equal

amount of A&T, G&C”

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Emmanuelle Charpentier/ Jennifer Doudna

2020

Genome Editing

CRISPR/Cas9 Gene Editing

sgRNA

DNA

Cas9

sgRNA

DNA

Cas9

sgRNA

DNA

Cas9

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DNA Sources

Eukaryotic Cells Prokaryotic Cells

Nucleus Mitochondria Chloroplasts

Cytosol

Eukaryote Mitochondria Nuclös

. 'O:uum

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Bacterial DNA

Staphylococcus pneumonia 2 strains: S type: form capsid, causes pneumonia.

R type: does not form capsid, cleared by

the immune system, does not causes pneumonia.

R

S

Survives

Pneumonia Smooth

Rough

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Frederick Griffith’s Experiment (1928)

R

S

S

R

S

Survives

Survives

Dies

Dies +

Heat- killed S

Smooth

Rough

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Bacterial DNA

R

S

Dies +

R

S

Dies +

+ Protease (cleaves proteins)

R

S

+

+ Nuclease (cleaves DNA)

Survives

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Bacterial DNA

R

S

S

R

DNA

Ø DNA of S type is transferred

to R type bacteria.

Ø The genetic material

transmitted is DNA.

+

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PART II: DNA Structure

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Inheritance in Genetics

Ø Traits are encoded by genes.

Ø Genes are found on DNA.

Ø DNA is organized into chromosomes.

Cell Nucleus Chromosome DNA Gene t of DNA}

Traits

https://www.civilsdaily.com/biotechnology-basics-of-cell-nucleus-chromosomes-dna-genes-etc/

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Structure of DNA

Ø DNA is a polymer of nucleotides (monomers). Ø Formed by 4 bases:

Purines: Adenine Guanine

Pyrimidines: Cytosine Thymine

Purines Adenine Deoxyribose Guanine Deoxyribose Thymine Deoxyribose Pyrimidines Cytosine Deoxyribose

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Nucleotide OH —P —O CH2 HOCH2 5 H 3 OH Nucleoside Base A, G, T, orc Position of HO 2 H 3 OH Base 2 H carbon atoms This group is OH in RNA. Phosphate Sugar Sugar This group is OH in RNA.

Structure of DNA

Nucleoside: Sugar + Base Nucleotide: Phosphate + Sugar + Base

Position of

carbon atoms

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5' end 5' end terminates with phosphate group 5' end Phosphate linked to 5' carbon and to 3' carbon Phosphodiester bonds 3' end 3' end terminates with hydroxyl (—OH) H 5'CH2 o H NH2 HO 3' end H 5'CH2 OH Figure 2.5 Three nucleotides at the 5' end of a single EX'lynucleotide strand. (A) The chemical structure of the sugar—phosphate linkages, showing the 5'-to-3 orientation of the strand (the red numbers are those assigned to the carbon atoms). (B) A common schematic Way to depict a cleotide strand.

Single Strand of DNA

A strand is a polynucleotide.

DNA Strand Formation:

The third carbon of the

deoxyribose sugar forms a phosphodiester bond with the phosphate group of the

following nucleotide.

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5' end HO 3' end

Single Strand of DNA

Formation of a

Phosphate/Sugar

backbone

Phosphate/Sugar

Backbone

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5' end 5' end terminates with phosphate group N H2 5' end Phosphate linked to 5' carbon and to 3' carbon Phosphodiester bonds 3' end 3' end terminates with hydroxyl (—OH) 5'CH2 0 5'CH2 5'CH2 HO 3' end Direction: NH2 Figure 2.5 Three nucleotides at the 5' end of a single polynucleotide strand. (A) The chemical structure of the sugar—phcxsphate linkages, showing the 5'-to-3' orientation of the strand (the red numbers are those assigned to the carbon atoms). (B) A common schematic way to depict a polynu- cleotide strand.

Single Strand of DNA

The DNA chain has an

orientation:

5’ end :

Phosphate Group

3’ end:

Hydroxyl Group

Direction:

5’ to 3’

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Formation of Double Stranded DNA

Ø DNA is formed by

complementary anti- parallel strands.

Ø Base pairing:

(A) Two hydrogen bonds attract A and T. —H —c —H Deoxyribose Adenine Deoxyribose H Guanine T Deoxyribose Thymine Three hydrogen bonds attract G and C. H N c H Deoxyribose Cytosine

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Formation of Double Stranded DNA

Strand#1:

5’ to 3’

5' end (terminates in S' phosphate) (6 HO S' end (terminates in 3' hydroxyl) S' end (terminates in 3' hydroxyl) S' end (terminates in 5' phosphate) OH p) Figure 2.8 A segment of a DNA molecule, showing the antiparallel orientation of the complementary strands. The overlying blue arrows indicate the 5'-to-3• direction of each strand. The phosphates (P) join the 3' carbon atom of one deoxyribose (horizontal line) to the 5' carbon atom of the adjacent deoxyribose.

Strand#2:

3’ to 5’

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Formation of Double Stranded DNA

Strand#1:

5’ to 3’

5' end (terminates in S' phos hate) S' end (terminates in 3' hydroxyl) S' end (terminates in 3' hydroxyl) OH o Figure 2.8 A segment of a DNA molecule, showing the antiparallel orientation of the complementary strands. The overlying blue arrows indicate the 5'-to-3• direction of each strand. The phosphates (P) join the 3' carbon atom of one deoxyribose (horizontal line) to the 5' carbon atom of the adjacent deoxyribose.

Strand#2:

3’ to 5’

Phosphate/Sugar

Backbone

Phosphate/Sugar

Backbone

Antiparallel

Strands

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of O Adenine Thymine Guanine Cytosine Guanine Cytosine Adenine Thymine Phosphate Deoxyribose sugar Base Oxygen Hydrogen Phosphorus C in sugar— phosphate chain Minor groove Major groove 34 Å per complete turn (10 base pairs per turn) D eter C and N in bases Figure 2.6 TWO representations Of DNA, illustrating the three-dimensional structure Of the double helix. (A) In a ribbon diagram. the sugar—phosphate backbones are depicted as bands. with horizon— tal lines used to represent the base pairs. (B) A computer model Of the B form Of a DNA molecule. The stick figures are the sugar—phosphate chains winding around outside the stacked base pairs,

Formation of the Double Helix

The two

complementary

anti-parallel

polynucleotide chain of

DNA are twisted

around one another to form a double helix.

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TRANSCRIPTION RNA PROCESSING NUCLEUS CYTOPLASM TRANSLATION Nudear envelope DNA mRNA Ribosome Potypeptide

Flow of the Genetic Information

TRANSCRIPTION CYTOPLASM TRANSLATION Ribosome Polypeptide

The Central Dogma

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Gene 3 DNA template strand mRNA TRANSLATION Protein A u c G c G A U A U A T u c G G c G G Gly G c A T U c c Gene 1 Gene 2 A A Codon Trp Amino acid Ser

The Genetic Code

Second mRNA base 8 0 UUIJ UUC I-JUA I-JUG CUIJ CUC CUA CUG AUC AUG GUIJ GUC GUA GUG Phe Leu Leu lie Met (M) or start Val (V) UCCJ UCC UCA IJCG CCU ccc CCA CCG ACU ACC ACG GCU GCC GCA GCG Ser pro (T) (A) UAU UAC UAA I-JAG CAU CAC CAA CAG AAC AAA AAG GAU GAC GAA GAG Tyr (Y) Stop Stop His (H) Gin Asn (N) Lys (K) Asp (D) Glu UGU Cys UGC UGA Stop UGG Trp(W) CGU CGC Arg (R) CGA CGG AGU Ser AGC AGA Arg AGG GGU GGC Gly (G) GGA GGG z

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PART III: Replication of DNA

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Why do we need to replicate the DNA?

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Why do we need to replicate the DNA?

Parental

Cell 2n

Daughter

Cells

2n

Cell Division

Without DNA Replication

Cell Division

After DNA Replication

Parental

Cell

Daughter

Cells

2n n

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Cell Cycle

Preparation to

division

INTERPHASE

G1

S G2

M

Cell Division MITOSIS

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DNA Quantity in the Cell During Cell Cycle

G1

S G2

M

DNA Synthesis

2n, 2C

2n, 4C 2n, 4C

2n, 4C

n= number of complete set of chromosomes

C= DNA content

in the cell

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DNA Quantity in a Cell

2n, 2C

2n, 4C

1n, 1C 1n, 1C

2n, 2C

2n, 4C

2n, 4C

2n, 2C 1n, 2C

1n, 1C 1n, 1C 1n, 1C 1n, 1C

Egg Sperm

Fertilization

S

G2

Mitosis Meiosis I

Meiosis II

G1

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DNA Replication

5' 3' 3' 5' (b) First, the two DNA strands are separated. Each parental strand can now serve as a template for a new, complementary strand.

5' 3' 3' 5' 5' 3' 3' 5' (c) Nucleotides complementary to the parental (dark blue) strand are connected to form the sugar-phosphate backbones of the new • daughter" (light blue) strands.

5' 3' 3' 5' (a) The parental molecule has two complemen- tary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C.

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First replication Second replication Parent cell (a) Conservative model. The two parental strands reassociate after acting as templates for new strands, thus restoring the parental double helix. (b) Semiconservative model. The two strands of the parental molecule separate, and each functions as a template for synthesis of a new, complementary strand. (c) Dispersive model. Each strand of both daughter molecules con- tains a mixture of old and newly synthesized DNA.

DNA Replication Model