Lecture 12: Genetic Recombination

Question 1

Why is genetic stability essential?

Answer 1

Genetic stability: essential to maintain genome, repair DNA, avoid cancer and avoid genetic disorders

Question 2

Why is genetic diversity essential?

Answer 2

Essential for meiosis and antibody diversity, also for evolution and adaptation of organism to new environment

Question 3

What are the 5 mechanisms of genetic recombination?

Answer 3

• homologous recombination
• non-homologous recombination
• transposition
• site-specific recombination
• independent assortment of chromosomes

Question 4

What is homologous recombination useful for?

Answer 4

Repair DNA damage, specially DNA double strand break

Answer 5

• end processing: after DNA break, DNA ends are processed by nuclease (RecBCD complex in ecoli)
• strand invasion: broken DNA invades an intact homologous DNA molecule, carried out by RecA
• intact DNA is used as template for replication to synthesise the missing DNA
• Holliday junctions are cleaved by resolvase RuvABC

Question 6

When the same Holliday junction is cleaved on both strands what is the result?

Answer 6

Non crossover DNA

Question 7

If different holiday junctions are cleaved on both strands what is the result?

Answer 7

Crossover DNA

Question 8

When does a cell carry out homologous recombination?

Answer 8

• DNA double stranded break
• end of double stranded linear fragment
• repair DNA single strand gap

Question 9

What is non homologous end joining useful for?

Answer 9

Repairing double strand DNA breaks during most of the eukaryotic cell cycle where there are no homologous chromosomes

Answer 10

• following double strand break, ends are recognised and protected by ku
• ku keeps ends next to eachother
• ends need to be processed, carried out by different enzymes to change DNA sequence and makes mutations
• DNA lig4 ligates 2 ends together

Question 11

What does recombinant do in site specific recombination?

Answer 11

Allows reciprocal exchange DNA stranded at specific target sites

-recognises its specific target and cleaves the double stranded molecule of DNA
- it exchanges DNA strands and relegates them, leading to genetic recombination

Question 12

What is the result of the site specific recombination when target sites are head to head orientation?

Answer 12

When target sites are inverted (head to head orientation) site specific recombination results in inversion of the sequence between them

Question 13

What is the result of the site specific recombination when target sites are head to tail orientation?

Answer 13

Site specific recombination results either in deletion OR integration

If process starts with one molecule containing both target sites, places target sites next to each other and exchanges DNA strands.

Results in 2 molecules: circular molecule containing composite site and original molecule without DNA sequence between target sites

Question 14

When circular e.coli molecules are linked, how do they separate?

Answer 14

• Ecoli has special head to head site in its terminus region (dif)
• site specific recombination at dif is carried out by XerCD recombinase
• results in formation of new crossover allowing separation of 2 linked chromosomes

Question 15

Bacteriophage lambda integration and excision

Answer 15

• following infection of e.coli host, bacteriophage can either carry out lytic or lysogenic cycle
• lambda phage circulars after entry into bacterial cell, and can insert into e.coli chromosome at att sites
• lambda phage uses Int protein to insert into chromosome
• composite sites that form are called attL and attR

Question 16

How are attP and artistes different?

Answer 16

attP in lambda contain several binding sites to accessory proteins
attB is much simpler

Answer 17

Target sites are head to head, alternates between 2 types of flagella

FljB encoded protein forming salmonella flagellum

FljA encodes regulator protein that represses expression of FljC which encodes different protein to form flagellum

Promoter for FljBA is between 2 head to head target sites. When promoter is infront of FljBA, salmonella is formed of fljB proteins

hin gene between 2 target sites encodes recombinase that can invert region by site specific recombination. When sequence is inverted fljAB not produced, so fljC is expressed

Question 18

Why is tail fibre variation useful in bacteriophage Mu?

Answer 18

Can express alternative tail fibres to expand its host range by being able to infect different hosts

Question 19

How is Mu tail fibre variation carried out?

Answer 19

• Mu tail fibre is made of 2 proteins
• first protein formed of 2 parts: a constant part encoded by the Sc gene and a variable part encoded by either Sv or Sv’
• second protein either encoded by U or U’
• gin gene decides recombinase that can invert portion of the genome between target sites