Topic 7.1: DNA Structure and Replication Flashcards Preview

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Flashcards in Topic 7.1: DNA Structure and Replication Deck (25)
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1

Harshey and Chase aim

Determine if DNA or proteins were the genetic material of a cell

2

Harshey and Chase experiment

1) It was previously known that viruses insert their genetic material into cells and so radioactively labeled viruses were prepared
2) Viruses grown in 35S had radioactive proteins but did not transfer this radioactivity to bacterium (remained in supernatant)
3) Viruses grown in 32P had radioactive DNA and did transfer this radioactivity to infected bacterium (found in pellet)

3

X-ray Diffraction by Franklin and Wilkins

1) X-rays will diffract when targeted at crystallised DNA molecules
2) The scattering pattern created can be used to determine structure

4

Deduced properties from X-ray diffraction

1) Composition: DNA is a double-stranded molecule
2) Orientation: The bases face inwards and the phosphates face out
3) Shape: DNA forms a double helix (10 bases per twist | 34 armstrong)

5

DNA Structure (4)

1) Phosphates form an outer backbone and nitrogenous bases are packaged within the interior
2) DNA is composed of an equal number of purines and pyrimidines
3) Two strands must run in antiparallel directions
4) A–T bond was the same length as a G–C bond

6

Deduction of DNA replication from DNA structure (2)

1) Replication occurs via complementary base pairing (A - T | G - C)
2) Replication is bidirectional (proceeds in opposite directions on the two strands) due to the antiparallel nature of the strands

7

Nucleosomes

1) DNA
2) 8 Histones
3) H1

8

Purpose of Nucleosomes

1) Makes DNA compact (better storage)
2) Prevents DNA damage (less exposed)
3) Assists in cell division (more mobility)
4) Involved in transcriptional regulation
5) Help to supercoil the DNA.

9

Eukaryotic Organization of DNA (5)

1) Nucleosomes are linked to form a string of chromatosomes
2) Chromatosomes coil to form a solenoid structure (~6 chromatosomes per turn)
3) Solenoid is condensed to form a 30 nm fibre
4) 30 nm fibres form loops, which are compressed and folded to form chromatin
5) Chromatin supercoils during cell division to form chromosomes

10

Non-Coding DNA

1) Satellite DNA (tandem repeats)
2) Telomeres (chromosome ends)
3) Introns (non-coding sequences)
4) Non-coding RNA genes
5) Gene regulatory sequences

11

Tandem repeats

Long stretches of DNA made up of repeating element

12

DNA Profiling

Technique by which individuals can be identified and compared via their respective DNA profiles

13

DNA Replication Enzymes

1) Helicase
2) DNA Gyrase
3) Single Stranded Binding Proteins
4) DNA Primase
5) DNA Polymerase III
6) DNA Polymerase I
7) DNA Ligase

14

Helicase (2)

1) Separates the DNA strands to form a replication fork
2) Breaks the hydrogen bonds between complementary base pairs

15

DNA Gyrase (2)

1) Reduces the torsional strain created by helicase
2) Prevents the DNA from supercoiling as it is being unwound

16

Single Stranded Binding (SSB) Proteins

Prevent DNA strands from reannealing

17

DNA Primase (2)

1) Generates a short RNA primer on each strand
2) Primers provide an initiation point for DNA polymerase III (DNA pol III can only add nucleotides to 3’-end of a primer)

18

DNA Polymerase III

1) Free nucleotides (dNTPs) line up opposite complementary bases
2) DNA polymerase III covalently joins free nucleotides together

19

DNA Polymerase I

Removes RNA primers and replaces them with DNA

20

DNA Ligase

Covalently joins the Okazaki fragments together

21

Okazaki Fragments

Discontinuous segments of DNA

22

Leading and lagging strand

1) On the leading strand, DNA pol is moving towards the replication fork and so can copy CONTINUOUSLY
2) On the lagging strand, DNA pol is moving away from the replication fork, meaning copying is DISCONTINUOUS.

23

DNA Sequencing

Process by which the base order of a nucleotide sequence is elucidated

24

Dideoxynucleotides

1) Lack the 3’-hydroxyl group necessary for forming a phosphodiester bond
2) Prevent further elongation of a nucleotide chain

25

DNA Sequencing Process

1) Four PCR mixtures are prepared – each with stocks of normal bases and one dideoxynucleotide (ddA, ddT, ddG, ddC)
2) Whenever the dideoxynucleotide is randomly incorporated, the DNA sequence is terminated at that base position
3) Because a complete PCR cycle generates millions of sequences, every base position is likely to have been terminated
4) These sequences are separated by gel electrophoresis to determine base sequence (according to ascending sequence length)
5) Automated machines can determine the sequence quickly if fluorescent labeling of the dideoxynucleotides has occurred