Genetic Variation

Question 1

Describe DNA Mismatch Repair

Answer 1

DNA repair is absolutely faithful (repaired strand matches original strand)
1. Excision
- damaged DNA base pair is cut out by series of nucleases

2. Resynthesis
   - repair DNA polymerase fills gap with a new DNA base pair
3. Ligation
   - DNA ligase seals nick left in sugar phoshphate backbone
   - Nick sealing, remakes broken phosphodiester bond between adjacent nucleotides

Question 2

Describe homologous Recombination

Answer 2

When does this occur?
shortly after DNA has been replicated but before cell has divided

1. Digest
   Nuclease digests 5’ ends of broken strands at the nick
2. Invade
   Enzymes allow for 3’ broken end to invade unbroken homologous DNA to search for complementary sequence via base pair
3. Elongate
   Repair DNA polymerase elongates invading strand (uses complementary strand as template)
4. Rejoin
   After repair polymerase passes point where break occured:
   - nwely elongated strands of rejoin og partner to form base pairs
5. Synthesis:
   Additional DNA synthesis at 3’ ends of both strands is followed up by ligation

DNA repair is reasonably faithful:
- closely matches original sequence on both strands
- might have slight differences (allelic differences = polymorphisms)

Question 3

Describe Non Homologous End Joining

Answer 3

1. Clean
   Nuclease chews back broken ends to produce flush ends (deletion of DNA sequence / loss of nucleotides)
2. Stitch
   flush ends stitched together by ligase

Unfaithful DNA repair

Question 4

Compare and Contrast the mechanisms of mutation within a gene and within regulatory DNA

Answer 4

Mutation within a gene:
- alter splicing of gene’s RNA transcript
- alter stability, activity, location , interaction of encoded protein / RNA product

Ex. Point mutation:
- caused by failures in copying and repairing DNA mechanisms
- low frequency of occurance (mismatch repair is effective 99% of the time)
- often has no effect on gene function (conservative sequence change does not affect amino acid sequence or occurs in non-essential portion of amino acid sequence or occurs in intron)
- rare to destroy gene activity; even more rare to improve gene activity

Regulatory Regions:
- can change expression of gene in population (if its advantageous)

ex. Lactose intolerance adults but now most of the world is lactose tolerant

Question 5

Describe what is an MEI and its function

Answer 5

Mobile Element Insertion

Transposition of MEI:
- specialized DNA sequences move from one chromosomal location to another
(recombines due to base pairing with similar short DNA target sequences present in both chromosomes)

• results in change in activity / regulation of gene
• promote gene duplication, exon shuffling and genome rearrangements
• alter expression pattern when inserted into regulatory DNA

Question 6

Describe the mechanism of Unequal Pseudo-homologous Recombination ‘Crossovers’

Answer 6

MEI allow homologous chromosomes to misalign –> formation of unequal template strands

• results in duplication of same gene on one chromosome
• duplicated genes undergo conservative mutational changes –> generates different but related genes (gene family)

Ex. Globin Gene family
1. ancestral globin gene was duplicated to create alpha and beta globin genes)
2. Further duplications generated sub family of beta globin (continues to branch out as more mutations arise)

Question 7

What happens during Exon Shuffling

Answer 7

Proteins are made up of domains.
Domains are coded by usually one exon.

Thus:
- exon shuffling and gene duplication create different combinations of exons
- each combination of exons codes for a unique protein with a distinct set of domains.

Diversity of human protein strucutre and function is generated by shuffling of exons in various combinations to generate unique functions for each gene

Transposable MEI drives Exon Shuffling
- two MEis of same type insert near each other in chromosome
- transposition mechanisms recognizes the ends of 2 diff elements and causes the DNA between the two MEIs to be excised into a new site.

Question 8

Describe how retroviruses interact with flow of genetic information

Answer 8

Retroviruses reverse normal flow of genetic information
- carry RNA as genetic material which they convert into DNA when infecting a host cell.
1. Reverse transcriptase enzyme (encoded by viral Genome and packaged with RNA) makes single-stranded DNA copy
2. creates second DNA strand to generate double-stranded DNA
3. DNA double heliz integrates with host chromosome –> allows for synthesis of viral RNA by host-cell RNA polymerase

Question 9

How to treat retroviruses

Answer 9

Block Viral Cycle

1. entry inhibitors (prevents ligand / receptor interaction )
2. Inhibitors of reverse transcriptase (prevents viral DNA production)
3. Integrase inhibitors (prevents viral RNA production)
4. Protease inhibitors (prevents viral protein maturation

Question 10

How do viruses reproduce

Answer 10

1. viral genome enters host cell
2. replicated to produce multiple copies –> transcribe and translated to produce viral coat protein
3. viral genomes assemble spontaneously with coat protein to create virus particles
4. particles escape cell via lysing (some DNA viruses remain latent for sometime)

Question 11

Why is it that closely related organisms have similar DNA sequences and genomes

Answer 11

Conservation of genetic information
- some DNA sequences that code for important cell functions and behaviours are conserved (ex. small ribosomal subunit RNA)

• exons are conserved the longest (across all species)
• intron conservation is also observed in mammals (species that are closely related to eachother have similar intron sequences)

Question 12

Vital statistics for human genome

Answer 12

• 21 000 protien coding genes
• 1.5% DNA sequence is exons
• 3.5% conserved with other mammals (including regulatory genes)
• single nucleotide alterations occur on average about 1 / 1000 nucleotides

Question 13

Explain steps involved in Sanger Sequencing and how to read a sequence off a gel

Answer 13

1) Primer
add primer to single stranded DNA
(primer sequence often determined by the restrictions where DNA is cut)

2) Chain Termination
dideoxy-nucleotides (no 3’ hydroxyl group) block DNA polymerase once incorporated
(DNA fragments are at diff lengths and ends with a specfic 3’ terminal dideoxy-nucleotides)

3) Electrophoresis
electrophoresis used to separate the differently sized DNA fragments. Fragments move towards positive end of electrophoretic gel + shorter fragments move fastest

Question 14

Explain Steps involved in Second/ Next Generation Sequencing
(how is diff from Sanger?)
(what is the importance of sequencing overlapping genomic fragments?)

Answer 14

1) Start
thousands of overlapping genomic fragments are placed on a spot on a micro-fabricated slide
- initiate reaction with primers (complementary sequence added in cloning step)

2) Stop + Pose
FLUOURESCENT reversible chain terminator dideoxy-nucleotides (dNTPs) are added
- can be digitally photographed once added to chain by DNA polymerase

3) Go again
fluorescent tags removed ( the nucleotide no longer terminates teh chain due to chemical modification)
- another fluroscent termination nucleotide is added to 3’ based on sequence

importance of overlapping sequenced gragments
- if cut in parallel –> requires lots of computing power)
- overlapping fragments can allow for more accurate and increased reads of each fragment –> greater depth)

Question 15

Identify what information about DNA is measured in Third Generation Sequencing.
How does this method differ from other strategies?

Answer 15

Examines intact DNA fragments directly
- DNA fragment is pulled through a nanopore and causes chanes in optial properties / ion fluxes

• reads both DNA sequence + nucleotide modification that alters transcription of coding regions

1. DNA passes through nanopore
2. Raw ouput: electrical signal caused by nucleotide blocking ion flow
3. each nucleotide has specific electric “signature”

(differs from other strategies because it doesn’t require DNA synthesis)

Question 16

Define ‘restriction (endo)nuclease’ and describe the relationship between the type, and frequency of ‘cuts’ to each enzyme based on the number of nucleotides recognized)

Answer 16

restriction endonuclease:
- enzymes that restrict transfer of DNA from one species of bacteria to another via cutting DNA into pieces
- used in genetic engineering:
- recognize specific internal DNA sequences (4 - 8 nucleotide base pairs)
- make reproducible cuts at the same restriction sites –> isolation of smaller DNA sequences used in DNA cloning

Different endonuclease:
(frequency)
- four base pair restriction nuclease cuts 1 / 256 nucleotides
- sic base pair restriction nuclease cuts 1 / 4096 nucleotides

(type)
- blunt ends
- 5’ overhangs –> used to facilitate joining of 2 diff pieces of DNA cut with same nuclease with high fidelity

Question 17

Define Cloning and Recombination

Answer 17

Cloning:
production of identical copies of DNA (amplification)

recombination:
cloned DNA is amplified in bacteria after being shuttle into bacterial plasmids

Question 18

describe steps in cloning a gene fragment from bacterial plasmids

Answer 18

Plamids consist of few thousand base pairs of circular DNA

linearized by restriction nuclease digestion + recombined with DNA sequence fragment to be cloned

1. Cut
   plasmid and DNA fragment to be cloned cut with the same restriction nuclease to have stick overhangs (used to base pair with eachother)
2. Paste
   After basepairing, plasmid DNA fragment is ligated together from 5’ to 3’ on each strand –> results in recombinant DNA
3. Amplified
   recombined plasmid DNA amplified in bacterial cultures:
   recombinant DNA is introduced into bacteria cell –> undergo cell division to clone the DNA of interest

Question 19

Compare + contrast ‘genomic’ and ‘cDNA’ libraries

Answer 19

genomic libraries:
- genomic DNA cut with restriction nuclease + all fragments recombined into plasmids (which are amplified in bacteria)

• specific DNA clones in library containing sequences are identified via hybridization probes

complementary cDNA libraries
- cDNA sequenced by RNA sequences
(changes to cells’ mRNA population due to development, physiological / disease can be assessed by generating cDNA libraries)
- complementary cDNA strand generated using reverse transcriptase (RNA –> DNA)
- og mRNA degraded via RNAse + second strand of cDNA generated using DNA polymerase

Both:
- cDNA’s can be cut with restriciton nucleases and made into libraries

Question 20

define hybridization probe + possible sources

Answer 20

DNA probe:
- single stranded DNA
- double stranded DNA is denatured (heat or acidity) to generate single strand DNA
- single strands when cooled + or at natural pH state will come together (renature) on basis of sequence-specific nucleotide base pairing
- if denatured single DNA strands are fixed onto surface : labelled complenetary single strand DNA probe used to identify specific DNA sequence / cone amongst large mixture of DNA sequences or clones via renaturation

possible sources:
- other genomes:
- RNA that has been reverse transcribed into DNA
- amino acid sequences in proteins
- databases

Question 21

explain steps involved in probing library to identify target gene sequences

Answer 21

1. place absorbent paper on top of a petrid dish with colonies of bacteria containing recombinant plasmids (library)
2. peel paper off from dish –> produces replica of colonies
3. bacteria lysed and DNA denature with alkali (DNA is bound to paper)
4. radioactively labeled DNA probe is added, incubated with colonies on paper and washed
5. Paper is exposed to photographic film (positions of desired colonies detected by autoradiography)

Question 22

explain how probes can be used ‘in situ’ to identify genes and transcripts in cells and tissues

Answer 22

labelled probes:
1. used to determine localization of DNA / RNA within tissues / cells (ex. identify virus-infected cells)

2. used to determine localization on chromosomes in cells / tissue
   - differently colored probes attach to genes on maternal and paternal copy of chromosome of interst in nromal cell (visualized at methaphase of mitosis after DNA replication)
3. used to identify chromosomal translocations in cells / tissues

Question 23

Explain sequential steps in each PCR cycle of amplifcation

Answer 23

Polymerase Chain Reaction: used to generate genomic or cDNA clones from complex mixture of nucleic acid fragments

Step 1.
add heat to separate double strand into single strands

step 2.
cool strands to anneal primers on each strand (cools to temp such that primers anneal first before the complementary strands)

step 3.
use DNA polymerase to initate DNA synthesis

step 1.
add heat to separate double strands…

1st cycle:
- amplification of two new DNA strands of DNA sequence are cloned (generates extended ends downstream of primers)
- intially, the clones are longer than the target sequence but each cycle slowly clones the right size

Each cycle produces an exponential amount of DNA clones:

After 3 cycles: the single DNA strands cloned are identical to the target sequence

Question 24

Define expression vector, recombinant protein and chimeric reporter genes

Answer 24

Expression vector:
- used to produce recomibnant proteins in bacteria, yeast, insect or mammalian cells
- specially designed vectors that induce transcription + translation signals to guide inserted gene that is meant to be expressed at high levels

recombinant protein:
- proteins expressed by recombinant DNA

chimeric reporter gene:
- gene encoding a protein whose activity is easy to monitor experimentally –> used to study the expression pattern of a target gene / localization of protein product

Question 25

describe 4 features of expression vectors

Answer 25

1. EGFP
   - gene that is tagged onto gene of interest in an in-frame ‘chimera’ used as a reporter

2 + 3. Upstream Cloning Site
and Downstream Cloning Site (multiple cloning sites)

• restriction enzyme cut sites allow gene of interest to be cloned directly upstream or downstream of the ECGP

4. Cytomegalovirus promoter
   - drives high expression of chimeric transgene after transduction of vector into mammalian cells

Question 26

Provide examples of medical uses for recombinant proteins + describe how transgene expression can be used to test gene function in cells

Answer 26

1. Medical use of recombinant proteins:
   - RNA- containing lipid nanoparticles used to deliver and express genes in cells including in vivo
2. transgene expression used to test gene function
   - modified expression vector used to produce integrin / EGFP chimeric reporter

a. EGFP tags integrin protein as green and integrin cDNA sequence generates focal contact protein
b. vector is transduced into migrating mesenchymal cells

Question 27

Explain the principles behind PCR-based site directed mutagenesis
(How are primers designed to generate substitutions + deletions)

Answer 27

Gene editing –> used to precisely alter gene sequences in test tube

‘wobbly’ primers during PCR-based amplifcation to introduce base pair changes, insertions, or deletions

Substitutions:
(extra primer bp)
- add new sequences to internal portion of primer not in target sequence

Insertions:
(extra primer bp @ end)
- used to add restriction site to end of target sequence to facilitate cloning
- can use internal nested primers to insert new sequences into middle of target sequence

Deletions:
(missing primer bp)
- removal of internal sequence from primer

Question 28

Explain components of the CRISPR / Cas9 system + how it is used to delete or insert genes / gene sequences into genome
(describe gRNA + Cas9 enzyme + how they work)
(discuss repair strategies + outcomes of double strand breaks that are generated by the Cas9 enzyme)

Answer 28

CRISPER system modified from bacteria where it recognizes specific viral DNA sequences and cuts them to degrade them

1. guide RNA:
   - contains self-pairing hairpin loop and 5’ terminal end of around 20 nucleotides that base pairs with any complementary target DNA
2. Cas9 enzyme:
   - binds to gRNA hairpin loop + guided to target DNA by 5’ end of gRNA via base pariing
   - short ‘PAM ‘ sequneces of 3 nucleotides are repeated every 8 to 12 base pairs in Human DNA
   - makes blunt double strand cut within target DNA sequence

Blunt doble strand cut:
- non-homologous end joining –> generates deletion often used to knockout gene of interest to study
- double strand breaks also used to insert sequences into cut via homogous recombination based repair (corrects deleterious genomic mutations)