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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

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Unit 1 Cell and Molecular Biology

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Unit 1 cell and molecular biology

Unit 1Cell and Molecular Biology

Section 8

The human genome project


History

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


Method

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


The human genome

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

?


Producing copies by pcr

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’


The polymerase chain reaction

The polymerase Chain Reaction


Understanding pcr

Understanding PCR

1.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 Taqpolymerase, isolated from thermophilic bacteria.

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

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.


Procedure

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.


Summary

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


Genome mapping

genetic map - allows relative positions

of heterozygous linked alleles to be located

(a)

I

I

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I

I

I

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I

I

I

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physical map - allows precise location of

specific DNA sequence to be located.

(b)

clone map

(c)

Sequence -Allows sequence of nucleotides to

be determined

AGGTCGCGATGCTA

Genome mapping


Genetic linkage mapping

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 recombinationfrequency.


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.


Physical mapping

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.


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.


Physical restriction mapping example

G

G

G

G

G

G

G

G

C

C

C

C

C

C

C

C

CUT FRAGMENT

Physical restriction mapping - example

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

  • 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


Unit 1 cell and molecular biology

C

G

G

G

G

G

G

C

C

C

C

C

G

G

G

G

G

G

G

G

G

G

G

G

C

C

C

C

C

C

C

C

C

C

C

C

G

G

G

G

G

G

C

C

C

C

C

C

Bam H1

EcoR1

EcoR1

plus

Pst1

Pst1

Bam H1

plus

EcoR1

Bam H1

plus

Pst1

Bam H1

plus

EcoR1

plus

Pst1

7

5

3

11

3

1

8

6

1

12

3

8

7

14

1

6

5

3

1

  • 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


Unit 1 cell and molecular biology

ANALYSING THE FRAGMENTS

15

12

9

3

6

0

kbp

A) BAM H1 (14 +1)

14

1

B) EcoR1 (12 +3)

12

3

Option 1

12

3

Option 2

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

3

11

1

Option 1

12

1

2

Option 2

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


Unit 1 cell and molecular biology

By repeating this procedure it is possible to build up a RESTRICTION MAP for this

section of DNA - this lets us know the base sequence at each point of cut,

15

12

9

3

6

0

kbp

A) BAM H1 (14 +1)

14

1

B) Pst 1 (8 +7)

8

7

Option 1

8

7

Option 2

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

8

6

1

Option 1

7

7

1

Option 2

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

Pst 1

EcoR1

BAM H1


Unit 1 cell and molecular biology

  • Checking the results of the triple digest EcoR1 plus Pst1 plus Bam H1 show that are

  • map is correct.

  • Since each cut with a known enzyme is a specific base sequence comparing restriction

  • maps 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 on

  • the DNA the more closely related the individuals

  • In the following example three endonucleases have been used (1-3) and have cut the strands

  • at the points shown. The reults indicate that individuals A and B are more closely related

  • that individuals A and C or B and C

1

1

1

1

2

3

A

1

2

3

B

1

2

3

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

C


Dna sequencing

DNA sequencing

1.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.


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.


Unit 1 cell and molecular biology

dideoxyribonucleoside triphosphate

deoxyribonucleoside triphosphate

Prevents strand extension at 3’ end

Allows strand extension at 3’ end


Unit 1 cell and molecular biology

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

Small amount of one

dideoxyribonucleoside

A

Rare incorporation of

dideoxyribonucleoside by DNA

polymerase blocks further growth

of the DNA molecule

OLIGONUCLEOTIDE primer

for DNA polymerase

5’

GCTACCTGCATGGA

CGATGGACGTACCTCTGAAGCG

3’

5’

single-stranded DNA molecule

to be sequenced

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


Separating dna fragments

DNA fragments can be separated by gel electrophoresis

Largest fragments

Smallest fragments

DNA moves to

the positive terminal

due to it’s overall

negative charge

+

gel with DNA

fragments

Separating DNA fragments:


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.


Automated dna sequencing

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

Automated DNA sequencing


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


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


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