Replication, maintenance and Rearrangement of Genomic DNA part 4

Question 1

Rearrangement of Immunoglobulin Light-chain genes

Answer 1

light chain genes are recombined and processed to generate a mature Ig mRNA

VJC segments

V=variable (250)

J=joining (4)

C=constant (1)

Internal rearrangement V-J

Transcription

Splicing

Translation

Question 2

What is the mechanism of VDJ recombination?

Answer 2

RAG1 and 2 are required to mediate VDJ recombination
RAG proteins recognize the RS (recombination signal) and bring both regions in close proximity

DNA cleaved (ds break) followed by re-ligation of coding strands

Final product: rearranged gene segment

VDJ recombination requires the proteins RAG1 and @ to mediate the process

Recognition of RS site (recognition signal)

The tow regions are brought to close poximity

The DNA is leaved (ds break) followed by religation of coding strand

Question 3

example of internal rearrangement V-J

VDJC segments

Answer 3

frist formation of D-J segments
Then V is joined to D-J (D=diversity)

Heavy chains are also undergoing rearrangement to form a mature mRNA

Transcription followed by splicing (RNA maturation)

Combination of different light and heavy chains can generate 20 x 10^6 different Ig’s.

Deletion or insertion mutation increase the potential number to 1011

Question 4

What is the structure of a T cell receptor

Answer 4

1. T cell receptors consist of 2 chains (alpha and beta) based on recombined V-J (alpha) or V-D-J (beta) added to a constant region fragment
2. role: The T cell receptor (TCR) is a complex of integral membrane proteins that participates in activation of T-cells activation in response to the presentation of antigen
3. the same type of rearrangements described previously is used to generate the T cell receptors

Question 5

Describe the steps of transposition via DNA/RNA intermediates: Bacterial Transposons

Answer 5

Transposable elements: first described by Barbara McClintock (in corn)

DNA or RNA mediated

DNA mediated contains insertion sequence coding for transposase sandwiched b 2 Inverted repeats (IR)

Transposition does not require replication of the transposon DNA

Transposase catalyzed the cleavage (blunt). The target site is staggered and will join the transposon. Gap filled by DNA polymerase and religated by DNA ligase

Cleavage (by transposase) blunt for the transposable element and staggered at the insertion point

Micked DNA unwound and transposon is inserted and ligated. Bap filled generating direct repeats at both ends

Question 6

Explain transposition via DNA/RNA intermediates: the organization of retroviral DNA

Answer 6

1. viral RNA

reverse transcriptase

Integrated proviral DNA flanked by LTR

Viral genes coded for Reverse transcriptase, polymerase (integrase) and structural proteins (env)

Question 7

Generation of LTR (long terminal repeats) during reverse transcription

Answer 7

1. LTR: 3 elements-r, U5, U3
2. R: short repeat sequence (20 nucleotides)
3. U5: unique to 5’ end of viral RNA
4. U3: unique to 3’ end of viral RNA

• initiation requires a tRNA primer binding to Priner Binding Site (PBS) common in plant viruses too
• RNA strand degraded (RNase H) Viral RT extends DNA from tRNA
• tRNA-U5-R DNA fragment will jump to other end of viral RNA. RT extends primer
• RNA-DNA hybrid degraded by RNase H
• Second strand of DNA synthesized. RNase H degrades tRNA. Second jump, hybridization to PBS and completion of DNA synthesis
• LTR is now U3-R-U5 at both ends

Question 8

structure of a LTR Retrotransposon

Answer 8

Yeast retrotransposon Ty1: high similarity with Retroviruses &elements are LTR.

In green are short repeats flanking the protein coding sequences and LTR’s

Similar types of retrotransposon are accounting for 8% of human genome

They code for their own Integrase and RT and can propagate themselves

Question 9

Structure of Human LINEs

Answer 9

1. LINE (interspersed: inserted at regular intervals) do not contain LTR
2. They possess A-tracts at 3’ end (An may have originated from RT of 3’ poly-A of mRNA)
3. They possess direct repeats

• staggared cleavage of target DNA
• tetrotransposon RNA used as intermediate
• RNA/DNA hybrid formation. after DNA synthesis the RNA is degraded by RNase H
• Second DNA strand synthesized

Question 10

Formation of a processed pseudogene

Answer 10

Formation of pseudogen involves RT

mRNA is poly-adenylated and spliced

Reverse-transcription + synthesis of complementary strand

Integration into chromosome

Pseudogenes do not possess the 5’ UTR (untranslated region, no promoter or regulatory elements). Once integrated into chromosome, they are not functional as genes.

Question 11

How are genes or DNA amplified?

Answer 11

1. Generation of multiple copies of a specific chromosomal region within a cell
2. Repeated rounds of replication
3. Thsi can increase the number of mRNA copies
4. Best known example: rRNA genes in frog eggs. They require large amounts of proteins to be synthesized
5. Gene amplification favors transcriptional increase of rRNA genes

Gene amplification also observed in cancer cells: resistance to methotrexate (inhibitor of DNA synthesis)

Gene amplification: multiple copies of specific chromosomal region.
Result: multiple copies of the same coding region within a cell (can amplify transcription)
rRNA in frog eggs undergo gene amplification (useful when large amounts of protein require to be synthesized).