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Unit 1 Cell and Molecular Biology Section 8 The human genome project

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Unit 1 Cell and Molecular Biology. Section 8 The human genome project. History. Late 1980’s idea was proposed Predicted it would take 15 years Cost about $200 million per year $1 per base pair Officially began in 1990 - PowerPoint PPT Presentation

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Page 1: Unit 1 Cell and Molecular Biology

Unit 1Cell and Molecular

Biology

Section 8

The human genome project

Page 2: Unit 1 Cell and Molecular Biology

History

Late 1980’s idea was proposed Predicted it would take 15 years Cost about $200 million per year

$1 per base pair

Officially began in 1990 26 June 2000 joint announcement from Blair and

Clinton ‘the draft complete’ Joint publication in Nature and Science 12 Feb 2001 14 Apr 2003 – The finished human genome

Page 3: Unit 1 Cell and Molecular Biology

Method

Genetic Mapping Identifies relative positions of genes E.g. Gene 2 lies between genes 1 and 3

Physical Mapping Absolute positions of genes on chromosomes E.g. Gene 2 is 1 million bp from gene 1

DNA sequencing Actual ATCG combinations

Page 4: Unit 1 Cell and Molecular Biology

The human genome

The total genetic complement of a cell

Has 3 billion (3 x 109) base pairs

Comprises 23 pairs of chromosomes

How did Biologists go about sequencing the genome ?

Page 5: Unit 1 Cell and Molecular Biology

Producing copies by PCR In order to sequence DNA it is necessary to produce

a huge number of exact copies of the original stands.

The technique used to do this is known as PCR or ‘Polymerase Chain Reaction’.

Once the copies of DNA have been produced they can be analysed.

Note – This is the technique used by forensics to amplify tiny samples of DNA for ‘fingerprinting’

Page 6: Unit 1 Cell and Molecular Biology

The polymerase Chain Reaction

5'3'

3 '5 '

5 '

3 '

3 '

5 '

5 '

3 '

3 '

5 '

Taq

Taq

4

8

16

etc.

DNA duplex (two strands)

primer

primer

denature

synthesis

repeat cycle

Page 7: Unit 1 Cell and Molecular Biology

Understanding PCR1. PCR can amplify any DNA sequence hundreds of millions of times in

just a few hours. It is especially useful because it is highly specific, easily automated and capable of amplifying minute amounts of sample

2. The whole process is only possible because of a special heat-stable enzyme called Taq polymerase, isolated from thermophilic bacteria.

3. The enzyme Tac polymerase is able to tolerate temperatures of 95C and has a temperature optimum of 72C.

4. This enzyme can synthesise the complementary strand of a given DNA strand in a mixture containing the four DNA nucleotide bases and two short DNA fragments called primers. Each primer is usually about 20 base pairs (bp) long. The primers are designed to bind to the DNA at either side of the target sequence.

Page 8: Unit 1 Cell and Molecular Biology

Procedure

Step 1 – The DNA is heated to 950 C breaking the hydrogen bonds and separating the strands.

Step 2 – The strands are cooled to between 55 – 700C and the primers added.

Step 3 – The strands are heated to between 70 – 72 0C so that Taq Polymerase can copy each strand from the point of the primer.

Page 9: Unit 1 Cell and Molecular Biology

Summary

PCR requires the following:-

Template DNA Primers – starting points for the construction of new

strands Taq Polymerase – a polymerase enzyme which

works at high temperatures Supply of nucleotides

PCR can amplify a single strand of DNA by a factor of millions

Page 10: Unit 1 Cell and Molecular Biology

Genome mapping

(a)

(b)

I II I I I I IIIIIIII I

I I I II I

(c)

AGGTCGCGATGCTA

genetic map - allows relative positionsof heterozygous linked alleles to be located

physical map - allows precise location of specific DNA sequence to be located.

clone map

Sequence -Allows sequence of nucleotides to be determined

Page 11: Unit 1 Cell and Molecular Biology

Genetic linkage mapping

1. Linkage mapping can be used to locate genes on particular chromosomes and establish the order of these genes and the approximate distances between them.

2. This idea is based on the fact that the further apart linked genes are on chromosomes the more likely crossing over will take place resulting in more recombinants being formed.

3. The greater the number of recombinants, the further apart linked genes are on a chromosome. Working out the number of recombinants relative to the parental genotypes gives a percentage value which is called the recombination frequency.

Page 12: Unit 1 Cell and Molecular Biology

Example:

Linked Gene Pair Recombination Frequency %

AB 11

AC 7

BC 18

The information from the table can be used to produce a chromosome map.

B A C

18 units

11 units 7 units

Page 13: Unit 1 Cell and Molecular Biology

Genetic Linkage Mapping relies on having genetic markers that are detectable – sometimes these are genes that cause disease, traced in families by pedigree analysis. The marker alleles must be heterozygous and be linked on the same chromosome so that recombination can be detected.

The overall result of genetic mapping is to produce a picture of the locations of the marker loci on the chromosomes – rather like establishing the order of the cities and large town between two points on a map.

Page 14: Unit 1 Cell and Molecular Biology

Physical Mapping

1. Physical mapping is required to add some more of the detail to what is obtained by genetic mapping

2. As with genetic maps, construction of a physical map requires markers that can be mapped to a precise location on the DNA sequence.

3. The distance between markers is usually expressed as a number of nucleotides in a physical map.

4. A physical map can be made by isolating DNA from a chromosome then cutting it using restriction enzymes (also known as restriction endonucleases) to construct a pattern.

Page 15: Unit 1 Cell and Molecular Biology

5. Different restriction enzymes cut the DNA at different points as each recognises a particular short sequence of bases occurring in the DNA. Where the sequence is recognised, the enzyme cuts the DNA so that it is cut into fragments.

6. By using combinations of restriction enzymes and working out the size of the fragments it is possible to recognise a pattern. The fragments can be identified by their size or by using a specific DNA probe to bind to its complementary sequence.

Page 16: Unit 1 Cell and Molecular Biology

Physical restriction mapping - example

1. Genes do not exist as separate entities but as part of the larger DNA molecule

2. DNA can be broken up into fragments by ENDONUCLEASES which cut at specific base sequences - look up page 316 of your textbook for more examples

G G C C

GGCC

CUT FRAGMENT G G C C

GGCC

Page 17: Unit 1 Cell and Molecular Biology

Not 1 - always cuts at the following sequence of eight pairs. This happens on average every 65536 nucleotide pairs ( 1 in 48 ) and produces much larger fragments

The resulting fragments can be separated using GEL ELECTROPHORESIS and used to physically map the sections of DNA - See textbook p317

The following table shows the fragments produced from a 15kbp fragment

Bam H1 EcoR1 Pst1 Bam H1 plus

EcoR1

Bam H1 plusPst1

EcoR1plusPst1

Bam H1plus

EcoR1plusPst1

141

123

87

1131

861

6531

753

G C G G C C G C

GCGGCCGC

G C G G C C G C

GCGGCCGC

C G C C G GGC

G C G G C C C G

Page 18: Unit 1 Cell and Molecular Biology

ANALYSING THE FRAGMENTS

0 3 6 9 12 15

kbp

A) BAM H1 (14 +1)

14

1B) EcoR1 (12 +3)

Option 1

Option 2

3

3

12

12

C) BAM H1 plus Eco R1 (11 +3 +1)

3 11

12

Option 2 for EcoR1 above would give 12 +2 +1 with the double digest which is INCORRECT. This means that the fragment must have been cut with the following orientation

EcoR1 BAM H1

1

Option 1

Option 2

1

2

Page 19: Unit 1 Cell and Molecular Biology

By repeating this procedure it is possible to build up a RESTRICTION MAP for thissection of DNA - this lets us know the base sequence at each point of cut,

0 3 6 9 12 15

A) BAM H1 (14 +1)

14

1B) Pst 1 (8 +7)

Option 1

Option 2

8

8

7

7

C) BAM H1 plus Pst1 (8 +6 +1)

Option 18 16

17

Option 2 for EcoR1 above would give 7 +7 +1 with the double digest which is INCORRECT. This means that the fragment must have been cut with the following orientation

7

EcoR1 BAM H1Pst 1

kbp

Option 2

Page 20: Unit 1 Cell and Molecular Biology

Checking the results of the triple digest EcoR1 plus Pst1 plus Bam H1 show that aremap is correct.

•Since each cut with a known enzyme is a specific base sequence comparing restrictionmaps allows biologists to look for the numbers and locations of these base sequences.The theory is that the greater the number of sequences and the closer their location onthe DNA the more closely related the individualsIn the following example three endonucleases have been used (1-3) and have cut the strandsat the points shown. The reults indicate that individuals A and B are more closely relatedthat individuals A and C or B and C

1 2 31

1 2 31

1 3 21

A

B

C

Physical Mapping relies on the availability of many copies of the DNA fragment. This is only possible because of the technique known as POLYMERASE CHAIN REACTION or PCR which allows many copies of the section of DNA to be produced

Page 21: Unit 1 Cell and Molecular Biology

DNA sequencing1. The final stage of the genome project is to determine and assemble

the actual DNA sequence itself.2. There are several critical requirements for this part:-

a. Single stranded DNA fragments must be generated as the templates;

b. sequencing technology must be accurate and fast;c. computer hardware and software must be available to analyse the sequence data.

3. The technique used for sequencing is called dideoxy chain-termination method.

4. This method relies on making a copy of the DNA template to be sequenced using:-a. a DNA polymerase;b. a primer;c. the four dNTPs (deoxyribonucleoside triphosphates dATP, dCTP, dTTP and dGTP) to extend the chain;d. a labelled dNTP; nowadays using a fluorescent dye rather

than a radioactive element as used in the past.

Page 22: Unit 1 Cell and Molecular Biology

1. In the correct conditions the polymerase can make a copy of the DNA by a process that is essentially the same as that used in DNA replication.

2. The chain termination part is what makes the key difference from normal DNA replication. This involves setting up four separate reactions, each including one of the four dideoxy NTPs (ddATP, ddGTP, ddCTP and ddTTP).

3. These modified nucleotides cannot form the next phosphodiester bond in the growing chain – hence when a ddNTP is incorporated into the copy, it terminates the process.

4. The large number of fragments that are produced in the four reactions produce a set of sequences that differ in length by one base, and end with a particular ddNTP.

Page 23: Unit 1 Cell and Molecular Biology

deoxyribonucleoside triphosphate

dideoxyribonucleoside triphosphate

Allows strand extension at 3’ end

Prevents strand extension at 3’ end

Page 24: Unit 1 Cell and Molecular Biology

5’

Normal deoxyribonucleoside triphosphate precursors (dATP, dCTP,dGTP, and dTTP)

Small amount of one dideoxyribonucleoside

Rare incorporation ofdideoxyribonucleoside by DNApolymerase blocks further growthof the DNA moleculeOLIGONUCLEOTIDE primer

for DNA polymerase

GCTACCTGCATGGACGATGGACGTACCTCTGAAGCG

3’single-stranded DNA moleculeto be sequenced

The fragments can be separated using gel electrophoresis -to see animated version of this go ‘DNA sequencing by enzyme methods’ click here

A

5’

Page 25: Unit 1 Cell and Molecular Biology

Separating DNA fragments:

DNA fragments can be separated by gel electrophoresis

DNA moves to the positive terminaldue to it’s overallnegative charge

fragments

+

Smallest fragments

Largest fragments

gel with DNA

Page 26: Unit 1 Cell and Molecular Biology

DNA sequencing:

separating the DNA fragments

The different fluorescent labels in the copied DNA strand are detected as they come off the bottom of the gel.

Page 27: Unit 1 Cell and Molecular Biology

Automated DNA sequencing

gel

A C T G A C

(a )

(b )

(c)

(d )

This gives a direct readout of the sequence and the process can be automated so that it is much faster than by conventional sequencing.

Page 28: Unit 1 Cell and Molecular Biology

To see an animated version of this process go to the ‘cycle sequencing’ section on the following website of the Biology Animation Library

http://www.dnalc.org/resources/BiologyAnimationLibrary.htm

Page 29: Unit 1 Cell and Molecular Biology

Activity

Look at the arrangements document to clarify what information is required.

Read DART pg 73 – 81.

Read the Monograph pg 67 – 79

Scholar – 8

Internet research