LSM1102 Notes

LSM1102 NotesThanks to Prof Wu Jinlu, Prof Cynthia He Yingxin and Prof Chew Fook Tim for their lectures and slides, of which these notes are based on.

Miscellaneous

Definition list

1. Centromerea. It is the DNA part of the chromatids that is below the kinetochore proteins, linking

sister chromatids2. Kinetochore

a. It is the protein found at the centromere, where spindle fibres attach during cell division

b. It has an inner plate that interacts with the centromere, and an outer plate which the positive end of the spindle fibre attaches

3. Centrosomea. It is an organelle in animal cells that act as the microtubule-organising centerb. Made from 2 centrioles

4. Centriolea. A structure made from 9 sets of triplet microtubules, which is made from tubulin

5. Chromosomea. It is a structure, made from DNA and proteins, containing genetic material of an

organismb. A pair of non-sister chromosomes are homologous, being nearly identical in shape

and sizec. Can be either chromatid or chromatin

i. Usually a pair of sister chromatids is called a chromosome during division.6. Chromatin

a. It is a chromosome during interphase, where it is loosely packed to allow for transcription.

7. Chromatidsa. It is a chromatin that has replicated. It has 2 forms: thread-like chromatids after S

phase, and condensed chromatids after prophaseb. It is used in the context of sister or non-sister chromatids of a chromosome during

division8. Tetrad/ bivalent

a. Both refer to the pair of homologous chromosomes associating with each other.b. However, tetrad can also mean when the 4 chromatids are clearly visible at

diplotene and diakinesis9. Horizontal gene transfer

a. When genetic material is transferred from one organism to another that is not its offspring

10. Conjugationa. It is the transfer of genetic material between bacterial through direct cell to cell

contact11. Conjugants

a. Cells that are undergoing conjugation, the DNA donor and recipient

12. Exconjugantsa. Any two cells that have previously mated

13. Transconjugantsa. The recombinant descendants of recipient exconjugants

14. Nucleotidea. The building block of DNA, consisting of a pentose sugar, a nitrogenous base and a

phosphate group15. Nucleosome

a. A subunit of chromatin, consisting of a histone octamer with DNA wrapped around it.

16. Genea. Controls the phenotype of an organism

17. Allelea. Alternate forms of a gene and is inherited from parents

18. Locusa. The site of a specific gene or DNA sequence on a chromosome

19. Genotypea. An organism’s collection of genes and their combination of alleles

20. Phenotypea. Physical expression of a genotype that may or may not be visible

i. Outward appearanceii. Proteins

iii. Behaviours21. Homozygote

a. A diploid organism having two identical alleles at the target locus on homologous chromosomes

22. Heterozygotea. A diploid organism having different alleles at the target locus on homologous

chromosomes23. Hemizygote

a. A diploid organism that has only one allele of a genei. Such as genes on the X-chromosome for males

24. Dominant allelea. An allele that is expressed when at least one copy is present

25. Recessive allelea. An alleles whose expression is masked by a dominant allele

26. Autosomea. A non-sex determining chromosome

27. Sex chromosomea. A chromosome containing genes that specify sex

28. Penetrancea. It is the proportion of a genotype that expresses a trait, even mildly

29. Expressivitya. It is the intensity of the phenotype of individuals of a particular genotype, out of

those already affected by that genotype30. Pleiotropy

a. A gene that causes multiple effectsb. Not all effects need be expressed in an individual

31. Multiple allelesa. When a given gene has 2 or more alleles that exist in the gene pool

Chromosomes and their transmission

Cell division

1. Cell cycle consists of interphase, mitosis/meiosis and cytokinesisa. Interphase has G1, S and G2 phase

i. Gap 1 phase is where the cell grows and functions normally. Nearing the S phase, the cell size doubles and more organelles are produced

Such as mitochondria and centrosomesii. Synthesis phase is where DNA duplicates, forming sister chromatids that are

still uncondensed.iii. Gap 2 phase is where the cells continues to grow and prepare for division.

2. Mitosis has 5 phases. Prophase, prometaphase, metaphase, anaphase, telophasea. It produces 2 genetically identical, diploid daughter cellsb. During prophase

i. Chromatin condenses into chromatids (or chromosome)ii. Nuclear envelope dissociates into smaller vesicles

iii. Mitotic spindle apparatus starts forming Consists of centrosome, spindle microtubules, and associated

proteinsc. During prometaphase

i. Centrosomes move to opposite ends of the cells, demarcating the two spindle poles

ii. Spindle fibres grow from the two poles, by rapid polymerisation of tubulin proteins

Astral microtubules grows in all direction Positions and orientates the spindle fibre apparatus

Polar microtubules grows towards the opposite pole When contacting the opposing polar microtubules. Contact

forces pushes the poles apart Kinetochore microtubules grows and attaches to kinetochore at the

centromere of sister chromatidsiii. The kinetochore on a pair of sister chromatids are attached to kinetochore

microtubules from both poles They attach to the outer plate of the kinetochore Failure to contact a kinetochore leads to depolymerisation and

retracts to the centrosome.d. During metaphase

i. Pairs of sister chromatids align themselves independently along the metaphase plate, moving by pushing and pulling forces

The kinetochore microtubules pulls the kinetochore towards the centrosome

The astral microtubules that contacts with the arm of the chromatid pushes it away from the centrosome

The balance of these two forces from both poles directs the chromosome towards the metaphase plate, where it oscillates awaiting anaphase.

ii. The attachment of kinetochore microtubules to the kinetochore is completee. During Anaphase

i. The kinetochore no longer joins the sister chromatids together at the centromere

Each chromatid is now only linked to one poleii. Anaphase A and B occurs concurrently, the centromeres move apart while

the microtubules shorten by degradation Anaphase A is the shortening of microtubules at the + end (the end

furthest from the centrosome) Either by ATP-driven motor protein causing both depolymerisation

and chromosome movement Or the microtubule disassembly drives chromosome movement

iii. Anaphase B is the growth of polar microtubules towards each other, pushing each other when the polar microtubules meet in the middle

f. During Telophasei. Chromatids reach respective poles and decondense into chromatin

ii. Nuclear membrane reforms around the chromatin, forming two separate nuclei

iii. Cytokinesis occurs concurrently Cleavage furrow forms and splits the parent cell into two

3. Prophase I of meiosis has several sub-stagesa. Leptotene is the condensation of chromatin threads to form chromosomesb. Zygotene is when homologous chromosomes pair up to form bivalents, with a

synaptonemal complex formingi. Homolog recognition occurs with a pairing centre, located at sister

chromatids at either the telomere or centromere. Proteins in the nuclear membrane recognise and bind to the highly

repeated sequence. The protein then moves around the membrane until it contacts the

other homologous pair A synaptonemal complex then forms between the bivalent

c. Pachytene is the crossing over between non-sister chromatids, forming a chiasma, with the synaptonemal complex fully formed

i. Synaptonemal complex causes the crossing overd. Diplotene is the start of dissociation of the synaptonemal complex, but

chromosomes are still linked by the chiasmae. Diaknesis is when the dissociation of the synaptonemal complex is finished, and the

dissolution of the nuclear membranei. After Diaknesis, the bivalent is still linked by the chiasma, and move together

to the metaphase plate, arranging themselves with independent assortment.

4. Prometaphase I occurs similarly to mitosis. Centrosomes move, 3 types of spindle fibre forms, attachment to kinetochore occurs

5. The rest of meiosis differs from mitosisa. During metaphase I

i. Pairs of sister chromatids are aligned along the metaphase plate in a double row instead of a single row

There is independent assortment of the two pairs in the double rowii. A pair of sister chromatids is linked to one of the poles, while the other

homologous pair is linked to the opposite pole.b. During anaphase I

i. The pair of homologous chromosome separates from each other, to opposite poles, while the sister chromatids of each chromosome are still connected

c. During Telophase Ii. The nuclear envelope forms around the sister chromatids at the poles

d. Meiosis II occurs similarly to mitosis, producing 4 genetically different haploid daughter cells

6. Hybrids can be infertile when they have an odd number of chromosomes, as during the meiosis checkpoint, the failure to find a homologous pair arrests the process of meiosis.

Bacterial diversity

1. Bacterial usually reproduce asexually by binary fission, thus horizontal gene transfer is needed to encourage genetic diversity

a. Conjugation involves direct cell to cell contactb. Transduction involves virusesc. Transformation involves uptake from the environment

2. During conjugation between a F+ cell and a F- cella. Cells that have conjugative plasmids (that contain the fertility factor) contain the

genes required for conjugationi. Such as for pilin protein needed for sex pilus formation

b. F+ cells form sex pili that initiates genetic transfer on contact with a recipient celli. They are hollow structures on the cell surface

c. On contact, relaxosome, a protein complex, separates DNA strands in the conjugative plasmid

i. Contains relaxase that cuts one strand of the plasmid DNA at the origin of transfer

ii. This produces a strand called T-DNA which gets transferredd. T-DNA-relaxase complex is recognised by the coupling factor which moves it into the

exporteri. Both the coupling factor and exporter are part of the sex pilus, and are

encoded by the conjugation plasmide. T-DNA-relaxase complex then moves through the mating channel into the recipient

celli. The other conjugation plasmid DNA strand remains in the donor cells

f. Relaxase then circularised the T-DNA in the recipient cellg. DNA replication processes in each cell then replicates the single stranded DNA into a

F plasmidi. Both cells are now F+

h. This process is similar even if the cell is F’i. F’ cells have chromosomal genes in their F plasmid

3. During conjugation between a Hfr cell and a F- cell

a. When a F+ cell has its F plasmid integrate into the bacterial chromosome, it becomes a Hfr strain

i. High frequency of recombinationii. This is because the F plasmid is an episome

It can replicate autonomously as a plasmid or be integrated into the chromosome.

b. The original of transfer determines the starting point and direction of the transferi. Chromosomal genes nearest to the origin are transmitted most efficiently

ii. The likelihood of transfer of genes decreases with distance from origin The transfer can break at any time, thus the probability of faraway

genes getting transferred is smallc. Transferred gene may not recombine with the recipient’s bacterial chromosomed. Unless the entire donor chromosome is transferred, the recipient cell does not

become Hfr too.i. Recombination of the F plasmid genes are also needed

4. Interrupted mating technique is a method to map the order of bacterial genes, based on their order of transfer into the recipient cell

a. A large number of Hfr and F- cells are mixed, allowing them to mateb. Conjugation can be interrupted at specific times using a blenderc. Cells are then transferred to various dishes with different media to test if certain

genes are transferred successfullyi. All the dishes must have a media that would exclude the growth of the Hfr

strain. There must be a selective marker for the two strains (usually resistance)

d. The time taken for the recipient cell to be able to grow in the specific medium will determine the order of the cells and their relative distances

i. To check for antibody resistant colonies, you need to replica plate an antibody-free medium plate to a plate with antibody

Done with sterile velvetii. If the timing is the same, the colony count will determine order

iii. To get distance, a graph can be plotted and the extrapolated backwardse. Minutes can be used as a unit of distance between genes, which the time it takes for

the gene to enter the recipient celli. The starting 0 minute point is assigned arbitrarily

f. As different Hfr strains have the fertility factor inserted into different point and directions, the whole chromosome can be mapped

g. The whole chromosome is usually not transferred due to several reasonsi. The mating channel is fragile and easily broken by environmental changes

ii. The time needed may be more than the average lifespan for the bacteriaiii. The recipient bacteria may lack the space for additional DNA

5. Transduction requires a bacteriophage to transfer DNA from one bacterium to anothera. Generalized transduction can occur with both temperate and virulent

bacteriophagesi. During the lytic cycle, the bacterial chromosome is fragmented and the virus

uses the cell’s machinery to manufacture more virusesii. A bacterial DNA fragment may be packaged into the viral capsid instead of

the viral DNA Any piece of chromosomal DNA may be incorporated.

b. Specialized transduction requires temperate bacteriophages that can undergo both lytic and lysogenic cycles

i. Specialized transduction only occurs if the lysogenic cycle happens, as a prophage must form.

A prophage forms when the viral DNA integrates into the bacterial chromosome

ii. When the lytic cycle occurs, the excised viral DNA may also contain a piece of bacterial DNA due to error in excision.

c. Lentiviruses, a form of retrovirus, is used as a vectori. Has a higher transduction rate as targets are specific due to surface

molecules on the virus recognising specific cells6. Genes that are fairly close together can be mapped by cotransduction, as more than 1 gene

may be packed into the phagea. The maximum distance depends on the maximum size of DNA that can fit into the

virusi. P1 can pack 2% to 2.5% of E.coli chromosome

ii. P22 can pack up to 1% of the Salmonella typhimurium chromosomeb. Plating twice is needed as once is to check that the reference gene is transduced,

and the second to check that the target gene is transduced as well.c. Ratio of the cell colonies on both plates will give the distance between the two

genesi. Colony count of cotransduced gene / Colony count of reference gene

ii. If grown on more than one plate, can do ratio of plates instead 1 plate testing for reference gene, 50 plates testing for target gene See how many out of the 50 plates have colony growth

7. Conjugation determines relative order and distance of genes, while cotransduction gives more accurate distances between the two genes

8. Transformation occurs when a bacterium uptakes extracellular DNAa. Natural transformation occurs without outside help

i. Only competent cells can undergo transformation, as they have genes coding for competence factors

The factors facilitate DNA binding to cell surface, uptake and integration

Expression of genes depend on environmental conditions b. Artificial transformation occurs with use of special techniques

9. All the above techniques involve replacing transfer where the DNA taken up undergoes homologous recombination with the homologous region of the DNA

a. Additive transfer can also occur where the incoming DNA is incorporated into the chromosome.

Chromosomes and their structure

1. Genome is all the chromosomes and DNA sequences an organism/ species CAN POSSESSa. It technically includes mitochondria DNA, but by convention it is not included unless

specified2. One nucleotide is 0.34nm long, and one turn of the DNA double helix is 3.4nm, containing 10

nucleotidesa. Diameter is 2nmb. DNA usually coils anticlockwise, in the right handed direction

i. It is called B-DNA3. For viruses

a. Their nucleic acid is surrounded by a capsid of proteinsi. RNA or DNA genome

b. They require hosts cells to replicate, and have a limited host rangei. Obligate parasites

c. Some viruses have a lipid bilayer taken from the host cell, containing proteins that aid in infecting another host cell

4. For bacteriaa. Bacteria have circular chromosomal DNA

i. A cell can have a single type of chromosome but with multiple copiesii. Only one origin of replication on each chromosome

iii. Repetitive sequences have a role in DNA folding, replication, gene regulation and genetic recombination

b. The bacterial chromosome is first packed with the formation of looped domains, then then supercoiling within the loops further compacts the DNA

i. The loop domains formation compacts about 10 timesii. The supercoiling further compacts it to 100 times

Overwinding makes the supercoil positive, as it twists the DNA in the direction of the double helix, which is anticlockwise like a right hand turning inwards.

This is for B-DNA (most common) Z-DNA is DNA twisted in the left-handed direction

Twisting a closed loop anticlockwise results in one region overlapping another heading in a direction to the top left

c. Supercoiling is regulated by two enzymes: DNA topoisomerase I and IIi. DNA topoisomerase I has a counter interaction with DNA topoisomerase II

DNA topoisomerase I increases the linking number by 1 making the supercoil more positive, without ATP

DNA topoisomerase II uses ATP and reduced the linking number of the supercoil by 2, such as from linking number of +1 to -1

DNA Gyrase is an example It breaks the double stranded DNA using ATP, then rejoins it

in a different orientation, reducing the linking number Most DNA are packed with negative supercoiling

Antibiotics can be drugs that inhibit topoisomerase, causing the chromosome to lose its structure, arresting processes like replication

Eukaryotic genome organisation

1. There are several difference between genomes in eukaryotic and prokaryotic cellsa. Eukaryotes have telomeres to protect the chromosome from degradation, and the 3’

end replication problemi. It acts as a biological ageing marker, and leads to senescence

b. Eukaryote genome is much larger with several origin of replication2. Chromosome organisation has 4 levels

a. Primary structure is like beads on a string, with nucleosomes linked by linker regionsi. Each nucleosome consists of a histone octamer

2 of each type of histone: H2A, H2B, H3, H4 Each histone has a N-terminal tail, and a C-terminal histone fold

N-terminal tails will determine the interaction between histones, and can be disrupted by modification like histone acetylation, methylation or promoted by other processes

The interaction between histones is like a histone code that controls gene transcription. It is affected by cell type and environment. It is hereditary giving rise to epigenetics

The histone contains many positively charged amino acids that bind to the phosphates along the DNA backbone

Lysine and arginine are the amino acidsii. The length is DNA around a histone is fixed at around 146/147 base pairs,

but the length of linker regions varies The diameter of the nucleosome is 11nm The DNA makes 1.65 negative superhelical turns around the core

iii. This structure was proven with digesting by DNase I Fragments produced where in multiples of 200bp Fragment consists of nucleosome DNA and linker DNA

b. Secondary structure is the organisation of nucleosomes to form a 30nm fibrei. This involves histone (H1) and non-histone proteins

H1 is less tightly bound to DNA, and helps compact adjacent nucleosomes.

ii. Salt concentration will affect the organisation as amino acids are charged and will interact with the ions present

At moderate salt concentration, H1 is removed, leading to primary structure

At low salt concentration, H1 remains bound and the nucleosome associate together into a more compact structure

iii. The nucleosomes zig-zag, and have little face to face contact with each otherc. Tertiary structure is radial loop model/ domain.

i. Formed from the interaction between the 30nm fibre and the nuclear matrix The Matrix-Attachment Regions (MAR) on the fibre gets anchored to

the nuclear matrix, creating radial loops The nuclear matrix consists of nuclear lamina and internal matrix

proteins Nuclear lamina is dense fibrillar network, which consists of

intermediate filaments (cytoskeleton) and membrane associated proteins

d. Quaternary structure occurs during metaphase, with the nuclear matrix being reorganised into a scaffold that the highly compacted radial loop can anchor to

i. Histone and non-histone proteins help to compact the DNA The compacted heterochromatin undergoes minimal gene

transcription ii. Both cohesin and condensin are multiprotein complexes that work like tongs

to push the DNA together compacting them. They have SMC proteins that use ATP to catalyses changes in

chromatin structure Structural Maintenance of Chromosomes

Condensin helps with chromosome condensation Condensin is usually in the cytoplasm and enters the nucleus

only during cell division It binds to chromosomes and compacts the radial loops,

changing the diameter but not loop number Cohesin helps with sister chromatid alignment

At the end of S phase, cohesin binds to uncondensed sister chromatids to hold them together

They remain bound until the middle of prophase, during which all the cohesins are released except those at the centromere

During anaphase, the cohesins at the centromere are released, allowing the chromatids to separate

Both differs from the synaptonemal complex that joins non-sister chromatids.

They differ in function and location3. During interphase, chromosomes have their own territory in the nucleus, a region of space

where they are usually founda. There are two levels of organisation of chromosomes during interphase

i. Euchromatin are less condensed regions of chromosomes This region is transcriptionally active Consists of 30nm fibres forming radial loop domains

ii. Heterochromatin is tightly compacted regions of chromosomes Generally, transcription does not occur at the heterochromatin

region, but it can occur at places like the telomeres and centromere The radial loop domains present are highly compacted

iii. There are also two types of heterochromatin Constitutive heterochromatin is always heterochromatic

Permanently transcriptionally inactive Usually contains highly repetitive sequences

Facultative heterochromatin can covert between euchromatin and heterochromatin

For example Barr bodies in females (the inactive X chromosome)

iv. By prophase, sister chromatids are entirely heterochromatin4. Chromosomes can be banded by different methods

a. Banded by supercoiling regions, positive and negativeb. Banded by packed-ness, heterochromatin and euchromatinc. Banded by staining of A-T rich regions

Molecular Genetics

Structure of DNA and RNA

1. Genetic material has several parts that meets its rolea. It contains the information necessary to code for an entire organismb. It is hereditable and is passed from parents to offspringc. It is replicable, to pass from within, and between organisms (hereditary)d. It is variable, to be able to react to changes

i. To account for the phenotypic variation seen between species2. The genetic material was first identified to be in the chromosome, which consists of both

proteins and DNAa. While DNA had only 4 nucleotides, ATCG, proteins have 20 different amino acidsb. There was more protein by weight that DNA in chromosomesc. Variations in proteins were identified, that they are responsible for different

functions3. Several experiments showed that the genetic material was DNA instead of proteins

a. Frederick Griffith experiments with Streptococcus pneumoniaei. Showed that genetic material survived heat treatment

ii. The bacterial cell type could change through the process of transformation Recessive, non-lethal type with heat-killed dominant, lethal type,

killed the lab ratb. Aver, MacLeod and McCarty experiments

i. Showed that the genetic material was DNA, and not RNA or proteinsii. Added DNA of dominant phenotype of recessive cell type and grew with

various enzymes None = grew

a. To show the extract contained genetic material DNase = no growth

a. To show DNA was responsible RNase = grew

a. To show RNA was not responsible Protease = grew

a. To show proteins was not responsiblec. Hershey and Chase experiment with Bacteriophage T2

i. Showed that genetic material was DNA not protein Bacteriophage T2 only has DNA and protein

ii. Either viral proteins or DNA were radioactively labelled with P or S To identify if the bacterial cell take up DNA or proteins

iii. Radioactivity of solution not taken up by bacterial showed more S than P present

Conclude that bacterial cells take up DNA rather than proteins. Thus DNA is the genetic material

4. Other research shows that RNA can also function as a genetic materiala. A.Gierer and G.Schramm showed that some viral genome contains only RNA, and

extracted RNA can still infect cells5. The basic unit of DNA is a nucleotide

a. It has a phosphate group

i. Bound to 5’ carbon of one nucleotide, and 3’ of anotherb. It has a 5 member sugar ring

i. 4 carbon and one oxygenii. Another carbon attached to the 4’ carbon, making this a pentose

iii. Difference between Deoxyribose and Ribose is the presence of a hydroxyl group on the ribose 2’ carbon

c. It has a nitrogenous basei. Purines are A,G, with 2 carbon rings joined together

ii. Pyrimidines are T,C and U with only one carbon ringiii. It is always attached to the 1’ carbon

d. A nucleoside is when the nitrogenous base binds with the pentose sugar, without any phosphate group

6. DNA is built in the 5’ to 3’ direction, using an existing strand, and an incoming nucleotide that is a triphosphate

a. The triphosphate group reacts with the 3’ carbon at the 3’ end, forming a phosphodiester bond, and producing pyrophosphate

i. Pyrophosphate (PPi) is two phosphate groups joined togetherb. The phosphate groups and pentose sugar forms the DNA backbone, with

nitrogenous bases projecting outwards in the same direction7. Rosalind Franklin’s X-ray diffraction pattern of DNA fibres suggested that DNA was helical

and contains more than one strand, with 10 bases per complete turn8. Erwin Chargaff showed that all four bases in DNA are in roughly equal amounts

a. Plus proportions of A=T, and as well as C=Gb. Became Chargaff’s rule

9. In a DNA double helixa. The two strands of DNA form a right-handed double helix

i. This B-DNA form is the predominant form in living cellsii. Under in-vitro conditions, A-DNA and Z-DNA can form

While A-DNA is right handed, it is more compact as the base pairs are not perpendicular to the helix-axis

Z-DNA has a left-handed double helical structureb. The bases in opposite strands form hydrogen bonds according to Chargaff’s rule

i. A with T forms 2 hydrogen bondsii. C with G forms 3 hydrogen bonds

c. The two strands are anti-parallel as they have opposite 5’ to 3’ directionalityd. There are 10 nucleotides per complete turn of the helix of 3.4nme. Has major and minor groves

i. Major groves are where the backbones are far apartii. Minor groves are where the backbones are close together

iii. The distance is the vertical line distance along the backbone, not the distance between base pairs, which is fixed

Due to the glycosidic bond between the nitrogenous base and pentose sugar being 120 degrees

120 degrees portion forms the minor groves The remaining 240 degrees forms the major groves

iv. The major and minor groves alternate, along the vertical line of the backbone

10. Most proteins binding occurs at the major groove

a. Binding can be nucleotide sequence specific, or non-specificb. The narrowness of the minor groves leads to the edges of the bases major grove

being more accessiblec. Proteins make contact with the sides of the bases at the major groove and interact

with themd. A synthetic DNA strand can also bind to the major groove to form a triple helix

i. This is a sequence specific binding11. DNA expression can be regulated by the expression of DNA mimic proteins

a. AbrB (repressor) has the complementary shape to DNA, and will bind to it to prevent transcription

b. AbbA (activator) has a shape similar to DNA, causing AbrB to bind to it instead, allowing transcription to occur

c. Sort of like competitive inhibition due to the shape of DNA grooves 12. RNA strands have a similar structure like DNA, but with a ribose sugar instead and U in place

of Ta. Complementary base pairing can still occur to give secondary structures

i. Bulge loopii. Internal loop

iii. Hair-pin loop These loops are formed from regions that are non-complementary,

and thus project out of the structureiv. RNA junction, like a 4-way street with a hole in the middle

b. Base stacking, involving VDW forces between nucleotides, also help to hold the secondary structure together

DNA replication and repair

1. A pulse-chase experiment has a labelled pulse, and the chase finds the prevalence of the label in each fraction/ generation

a. Proves that DNA replication is semi-conservative, with the original pulse as the parent strands being made from Nitrogen-15

i. The label is the heavier isotope of Nitrogen that is examined through equilibrium density-gradient centrifugation

Separates DNA molecules containing denser strands into its own band

2. DNA replication occurs with several proteins, in the 5’ to 3’ directiona. Helicase first breaks the strands apart at the origin of replication (oriC), creating a

replication forki. Prokaryotes only have one oriC on their circular DNA chromosome

DNA synthesis occurs bidirectionally from the origin of Chromosomal replication

Replication ends when the two replication forks meet, at the ter sequence, separating the two chromosomes

ii. Eukaryotes have multiple oriC due to their length DNA replication still processes bidirectionally from each oriC Eventually all the replication forks merge to give two chromatids

b. The single strands are kept apart by proteinsi. Prokaryotes have single-stranded binding proteins

c. RNA primase creates RNA primers, allowing replication in the 5’ to 3’ directioni. Only one needs to be created at the origin of replication on the leading

strandii. Lagging strand requires continues synthesis of a RNAprimer

d. DNA polymerase synthesises a daughter strand, in the 5’ to 3’ direction, using the parent strand as a template, through complementary base pairing

i. Prokaryotes have DNA polymerase I to V I and III are involved in normal replication

I is a single polypeptide and removes RNA primers and replaces them with DNA

5’ to 3’ exonuclease digest the DNASimultaneous 5’ to 3’ polymerase activity to replace

with DNA III has 10 different subunits and is the main replicating DNA

polymeraseThe alpha subunit in the DNA polymerase III

holoenzyme synthesises the DNAThe epsilon subunit does the proofreadingThe theta subunits stimulates the proofreading

II, IV and V are used in DNA repair and replication of damaged DNAe. DNA ligase covalently links DNA okazaki fragments together, removing the nicks

i. Uses energy from ATP or NADf. Topoisomerase (DNA gyrase) removes the supercoiling caused by helicase

i. As helicase moves down the strand, there is positive supercoiling as the double helix is broken into linear strands.

ii. DNA gyrase uses energy from ATP to alleviate the supercoilg. Topoisomerase IV also helps separate the two chromosomes after the termination

of DNA replicationi. The two chromosomes are intertwined, forming a catenanes

3. Bacterial oriC has DNA sequences needed for its functiona. AT-rich regions

i. Strands are easier to separate as it has fewer hydrogen bonds compared to CG

ii. Located next to the DnaA boxesb. DnaA boxes

i. Several are found at the oriCii. DnaA proteins bind here to initiate replication

iii. They form a large complex that spans across all the boxes, causing the DNA region to wrap around

Mechanical stress separates the nearby AT-rich regioniv. Allows helicase to bind to the oriC

c. GATC methylation sitesi. Regulates rates of replication and DNA replication mismatch repair

ii. Parent strands have the adenosine methylated, while the daughter strand does not

Allows differentiation of strands to know the template for mismatch repair during replication

iii. Another round of replication can only occur when both strands have their GATC sequence methylated

Prevents concurrent replications4. Opposite of the oriC is a pair of termination sequences (ter) on the DNA

a. The protein termination utilization substance (tus) binds to these sequences to stop the movement of replication forks

b. They physically prevent further movement of helicase, which would otherwise unwind the daughter from the parent strand

i. They can only block movement from one direction, thus you need two in total

If the helicase comes from the other direction, it is allows to pass to be stopped by the other tus protein

ii. As both tus protein blocks movement from coming from the same side, the region between the tus proteins is only separated once

5. As eukaryotes have linear DNA, they have telomeres at both endsa. Telomere refers to the complex of telomeric DNA sequences, and the associated

bound proteinsi. Each telomeric DNA sequence contains multiple tandem repeats

ii. They also have a 3’ end overhang that is 12-16 nucleotides long Due to 3’ end replication problem

b. Telomeres shorten with each round of replicationi. Telomerase can lengthen the telomeres, preventing their loss

Contains proteins and RNA The RNA is complementary to the repeating sequence in the

telomeric DNA sequence Allows binding to the 3’ end overhang

ii. Only the 3’ end strand is lengthened by telomerase Binding occurs through complementary base pairing Extension occurs through reverse transcription of RNA in telomerase Translocation occurs to allow the telomeric DNA to be further

extendediii. The other 5’ end is lengthened by normal DNA replication mechanisms

Primer, DNA polymerase, ligase6. There is a low rate of mutation due to several factors

a. DNA polymerase has proofreading mechanismsi. Configuration of DNA polymerase active site makes it unlikely to catalyse

bond formation between mismatched pairsii. Epsilon and Sigma subunits in prokaryotic DNA polymerase III allow for

proofreading, as mismatched pairs are unstable 3’ to 5’ exonuclease activity removes the incorrect nucleotide

Unlike the 5’ to 3’ exonuclease in DNA poly I The DNA polymerase II then moves back and resumes DNA synthesis

in the 5’ to 3’ direction b. DNA repair mechanisms

i. There is instability of mismatched pairs, leading to recognition by repair mechanisms

Mutation caused by change of nitrogenous base, or deletion of base Cytosine can be deaminated into uracil, changing the

complementary base pair from guanine to adenine Nitrogenous bases can also dimerise which leads to deletion or

change in base of the daughter strand ii. Base excision repair removes a single mismatched base pair

DNA glycosylase breaks the glycosidic bond between the pentose sugar and nitrogenous base, removing the nitrogenous base

Apurinic/apyrimidinic (AP) endonuclease and phosphodiesterase removes the associated sugar phosphate

The single nucleotide gap is then filled by DNA polymerase and DNA ligase

iii. Nucleotide excision repair removes a section containing a nitrogenous base dimer

Both eukaryotes and prokaryotes use DNA excision nuclease and DNA helicase

In prokatyotes, the DNA excision nuclease first cuts the sugar phosphate backbone around the damage, then DNA helicase removes the damaged portion

In eukaryotes, DNA helicase first unwinds the DNA locally, followed by DNA excision nuclease freeing the damaged portion

Prokaryotes remove around a 12 nucleotide portion, eukaryotes remove around a 30 nucleotide portion

The gap is then filled by DNA polymerase and DNA ligaseiv. Double-strand breaks repair joins blunt ends together

Nonhomologous end joining

The DNA portion at the site of the break is damaged and is degraded

DNA ligase then seal up the gap, leading to a deletion of bases

Homologous recombination Unlike nonhomologous end joining, this restores the original

DNA sequence by using the sister chromatid as a template Only occurs when the damage happens after DNA

replication, but before cell division occurs Is also used to join broken ends during chromosome

crossovers in meiosisc. Coordination between DNA replication and cell division

i. Ensures DNA replication does not happen excessively, which would increase the chances for mutation

ii. Controlled by availability of DnaA proteins manufactured by the cell As the number of DnaA boxes doubles after replication, there is

insufficient DnaA proteins needed to initiate another round of replication

iii. Also controlled by GATC methylation sites DNA adenine methyltransferase takes time to methylate the

adenine at the GATC methylation site on the daughter strand Having only one strand methylated does not efficiently initiate

replication

From DNA to protein

1. A gene is a transcriptional unit, and gets transcribed into RNAa. Structural genes encodes for a polypeptide, and produces mRNA when transcribed.b. Nonstructural genes do not result in a protein product, as the RNA transcript is the

final producti. Such as tRNA, rRNA, microRNA

ii. They have important cellular functioniii. They can either work along, or be part of a protein complex with protein

subunits2. Transcription takes place along the template strand, producing a RNA transcript that is

similar to the coding/sense stranda. RNA has Uracil instead of Thymine

3. During transcription in prokaryotesa. Initiation

i. Transcription factors bind to the gene promoters There is a consensus sequence gathered from all the genes in the

organism. The closer the promoter sequence resembles this, the greater the

rate of transcriptionii. The transcription factors recruits RNA polymerase to bind to the promoter,

forming a closed promoter complex RNA polymerase holoenzyme consists of 6 subunits

A core enzyme of α 2ββ ' ω

A sigma factor of σ The sigma factor recognises and binds to the DNA

It recognises the -10 (Pribnow box)and -35 sequences RNA synthesis occurs in the 5’ to 3’ direction

iii. The RNA polymerase then breaks the hydrogen bonds in DNA, forming an open promoter complex

b. Elongationi. RNA polymerase slides along the DNA in an open complex to synthesise the

RNA transcriptii. The first amino acid (N-terminal) is always methionine

c. Terminationi. Eventually a termination signal is reached, causing RNA polymerase to

dissociate from the DNA4. As prokaryotes do not have a nucleus, the RNA transcript is exposed to the cytoplasm as it is

transcribed, allowing translation to take place concurrentlya. Coupling of the two process occurs as the ribosome will attach to the RNA 5’ end

once it is long enoughb. A polyribosome forms as the mRNA transcript has multiple ribosomes bound to it,

translating it5. Translation occurs only with the mRNA produced from transcribed structural genes

a. The genetic code is written in codons of 3 nucleotidesi. The codons are degenerate as amino acids are coded for by more than one

codonii. The first two nucleotides determine the amino acid in most cases, thus the

third nucleotide can varyiii. Some codons are translated faster than others, between similar and

different amino acidsb. Requires tRNA, mRNA and rRNA with ribosomes

i. tRNA is able to recognise a triplet codon in mRNA, and carries the corresponding amino acid

They recognise using an antisense RNA sequence to the codon Aminoacyl-tRNA synthetases catalyses the binding of a specific

amino acid, to their appropriate tRNA The amino acid is attached to the 3’ end of tRNA via an ester

bond 20 types of synthetases for the 20 types of amino acids

ii. The ribosome complex consists of a large and small subunit Svedberg units is a measure of their rate of sedimentation In prokaryotes, the 30S and 50S subunits forms a 70S ribosome

complex The 30S subunit binds to the Shine-Dalgarno sequence on

the mRNA, through complementary base pairing with the 16S rRNA in the subunit

In eukaryotes, the 40S and 60S subunits forms a 80s ribosome complex

iii. The ribosome complex has three sites for tRNA ‘E’ site for the exit of tRNA ‘P’ site holding a tRNA with the polypeptide chain

Peptidyltransferase catalyses the formation of peptide bond ‘A’ site where a aminoacyl-tRNA enters and binds to the mRNA

iv. The first amino acid is always a modified form of methionine N-Formylmethionine

c. Termination occurs when the ribosome’s ‘A’ site reaches a stop codon, UAG UAA or UGA

i. There are no tRNA complementary to themii. Release factors will enter the ‘A’ site to dissociate the ribosome complex

from mRNA The structure of the release factor is similar to the shape of tRNA

6. Transcription in eukaryotesa. Unlike prokaryotes, eukaryotes have three RNA polymerase instead of one

i. RNA polymerase I transcribes all rRNA genesii. RNA polymerase II transcribes all structural genes and some snRNA genes

Thus it produces all the mRNA in the celliii. RNA polymerase III transcribes all tRNA genes and the 5S rRNAiv. However, they share a structural similarity and consists of subunits

b. The RNA polymerase binds to the promoter which has consensus sequencesi. The promoter contains at least 2 out of 4 elements

BRE element is located at -35 and recruits TFIIB TATA element is located at -30 and recruits TBP INR element is located at +1 and recruits TFIID DPE element is located at +30 and recruits TFIID

ii. There is a basal level of transcription when RNA polymerase binds to the core promoter

Additional regulatory elements either simulate or inhibit transcription

Enhancers recruits activators that hasten the assembly of the complex

Silencers recruits repressors that inhibit the assembly of the complex

iii. While prokaryotes only need the sigma factor in RNA polymerase, eukaryotes form a complex with the RNA polymerase

General transcription factors give the basal level of transcription Histone-modifying enzyme loosens the DNA from the histones to

allow easier access to the template strand Chromatin remodelling complex reverts the supercoiling caused by

the unwinding of DNA by RNA polymerasec. As there is a nucleus to separate the nucleoplasm from the cytoplasm, RNA

processing can take placei. 5’ capping and 3’ polyA tailing helps stabilize the mRNA and its translation

Different enzymes complexes add the cap and tailii. RNA splicing to remove introns and connect exons

Alternative RNA splicing give rise to variation is proteins coded for by the same gene

Spliceosomes are a complex of snRNA and proteins They are flexible to allows alternative splicing

iii. Defective mRNAs will be degraded by exosomes within the nucleus

Only fully processed mRNA can be exported through the nuclear pore into the cytoplasm

7. Translation in eukaryotesa. Initiation

i. Starts from the first AUG codon after the 5’ cap Both 5’ cap and 5’ polyA tail recruits initiation factors (eIFs) for

assembly of tRNA-mRNA-ribosome complex The AUG codon follows Kozak’s rule

An A or G at the -3 position One G at the +4 position, immediately after the +1 to +3

AUGii. Smaller, 40S subunit binds first to the mRNA, using 18S rRNA

60S subunit associates with the 40S to form 80S ribosome complex The first aminoacyl-tRNA complex is recruited to the ‘A’ site

Carries methionine instead of f-methionineb. Elongation

i. Occurs similarly to prokaryotes Catalysis of peptide bond followed by translocation

c. Terminationi. Unlike prokaryotes that use several types of release factors, eukaryotes use

one release factor, eRF1, to recognise all three stop codons8. Gene expression is regulated by all the processes between DNA to protein

a. Transcriptioni. Rate of assembly of RNA polymerase complex

b. RNA processingi. Effectiveness of modification

ii. Alternative RNA splicingc. Exporting

i. Efficiency of export proteinsd. Translation

i. Rate of ribosome complex assemblyii. Structure of mRNA

iii. Rate of degradation of mRNAe. Protein modification

i. Molecular chaperones to help protein foldingii. Covalent modification needed for proper protein folding

Glycosylation Phosphorylation Acetylation

iii. Cofactors and other protein subunitsiv. Degradation by proteasome, especially for incorrectly folded proteins

Methods to study gene expression and function

1. Classic genetics studies how the phenotype of an organism arises from its genesa. First the genome of the organism is exposed to mutagens

i. Chemicalsii. Irradiation through UV or X-rays

iii. Transposons that insert randomly into the genome

b. Then the offsprings are screened for phenotypical changesi. Limitation is that only specific and distinct abnormalities can be detected

c. Mutated offsprings are identified and the general location of the mutation is identified

i. Through finding the recombination frequency of the mutation, using known gene loci as references

d. The mutated region is then cloned and sequenced2. Reverse genetics studies how a gene can affect the phenotype

a. Using a gene of known sequence, all functions that requires it are identifiedb. The gene of interest or its expression is manipulated

i. Deleting the geneii. Changing the expression pattern of the gene

Expression levels Expression location Expression time in stage of development

iii. Tagging with a reporter Green Fluorescence Protein

iv. Genome editing of a specific portion of the gene Uses the repair mechanism for double stranded breaks to introduce

new genes or delete targeted ones Nonhomologous repair to delete nucleotides around the

break Homology-directed repair to add genes by introducing a

template with the new geneIn lab we introduced with heat shockTemplate can also be introduced with

microinjectionGene coupled with reporter to indicate successful

uptake into the genome Breaks in the original genome can be directed

Zinc finger nucleasesZinc finger binds to specific nucleotide sequencesPositions DNA endonuclease at target siteOne required for each strand of DNA

Transcription activator-like effector nucleases (TALENs)A protein with each di-peptide being able to bind to

a specific nucleotidePositions DNA endonuclease at target siteOne required for each strand of DNA

CRISPR CAS9Uses bacterial CAS9 to target specific locations in

the genomeTargeting directed by supplied guide RNAGuide RNA forms a complex with CAS9, which can

cause a double stranded breakc. The phenotypical changes are then noted and compared with different gene

manipulations3. Green Fluorescence Protein is a good reporter due to its properties

a. It requires no cofactor, so it always gives a signal when translatedb. GFP has been modified from its wild type sequence to make it more efficient in

mammalsi. A Valine codon has been added after the starting methionine to fulfil Kozak’s

ruleii. Changed the codons of existing amino acids to their more expressed form

Population genetics

Inheritance types

1. Factors influencing phenotypic ratiosa. Complete or incomplete or co-dominance or sex-influenced or sex-limited

i. Partial dominance leads to intermediate phenotypesii. Sex-influenced dominance will cause intermediate to have different

phenotypes based on genderiii. Sex-limited will only have the phenotype expressed in one gender

b. Autosomal or sex-linkedi. More males are affected when sex-linked

c. Monohybrid or dihybridd. Lethal or non-lethal

2. Expected phenotypic ratios in F2, with opposite homozygous parents (heterozygous F1)a. Autosomal complete dominance monohybrid cross

i. 3 dominant: 1 recessive b. Autosomal incomplete dominance monohybrid cross

i. 1 dominant: 2 intermediate: 1 recessivec. Autosomal complete dominance dihybrid cross

i. 9 dominant-dominant: 3 dominant-recessive: 3 recessive-dominant : 1 recessive-recessive

d. X-linked dominanti. Male:

ii. Female:e. X-linked recessive

i. Male:ii. Female

f. Y-linkedg. Recessive epistasis of autosomal complete dominance dihybrid cross

i. 9 dominant-dominant: 3 dominant-recessive : 4 recessive-_ Because recessive-dominant looks like recessive-recessive

h. Dominant epistasis of autosomal complete dominance dihybrid crossi. 12 dominant-_: 3 recessive-dominant : 1 recessive-recessive

Because dominant-recessive looks like dominant-dominanti. Complementary loci of autosomal complete dominance dihybrid cross

i. 9 dominant-dominant : 7 recessive Because any recessive will give the same phenotype

j. Duplicate dominant genes of complete dominance dihybrid crossi. 15 dominant : 1 recessive-recessive

Because any dominant will give the same phenotypek. Duplicate genes with cumulative effect

i. 9 dominant-dominant : 6 dominant + recessive: 1 recessive-recessivel. Dominant and recessive interaction

i. 13 dominant-_ : 3 recessive-dominant Because dominant-_ looks like recessive-recessive

3. Reading pedigreea. Determining type of inheritance

i. Autosomal dominant diseases Affects both males and females equally Does not skip generations

ii. Autosomal recessive diseases Affects both males and females equally Does skip generations

iii. X-linked dominant More affected females than males

May be lethal in hemizygous males, causing decreased male count

Affected father always gives affected daughter Does not skip generations

iv. X-linked recessive More affected males than females Affected mother always produces affected sons May skip generation

v. Y-linked Only males are affected Affected father always produces affected sons

b. Tipsi. If a generation is skipped, the inheritance is definitely recessive

Mendelian Genetics

1. There are several reasons why Mendel used the pea plant for this experimentsa. It has well defined characteristicsb. Pure-bred varieties are availablec. The flowers have male and females parts, allowing for self-fertilisationd. Both self and cross fertilisation are possiblee. Relatively short life cyclef. Large number of offspring produced from each mating

2. Mendel’s law of segregation states that the gamete only contains one allele of a genea. Thus both alleles of a parent’s gene are not inherited together in one gamete

3. Mendel’s law of independent assortment states that each allele in the parental pair, will combine randomly with either allele in the other parental pair, when forming the offspring

4. Partial dominance occurs when each genotype has a distinguishable phenotype, with the heterozygote having an intermediate trait between the two homozygotes extremes

5. When considering n numbers of locia. 2n types of gametes will be producedb. 3n types of offspring genotypes will be produced

c. 2n types of offspring phenotypes will be produced when there is complete dominance

i. Else there will be 3n types of phenotypes if all are incomplete dominance6. Testcross is used to identify the genotype of an organism, whether it is heterozygous or

homozygous for the genea. The organism is mated with a homozygous recessive tester stockb. If only one offspring phenotype is observed, the original organism is homozygous.

7. Reciprocal cross is used to determine if the gene is sex-linkeda. A homozygous dominant male is mated with a homozygous recessive female and

vice-versab. Autosomal genes will have equal results, while sex-linked genes will have different

ones8. Males are more affected by recessive X-chromosome diseases, as they are hemizygous9. There is a criss-crossing of recessive X-chromosome alleles

a. Affected mothers will definitely give affected sonsb. Affected fathers will definitely give carrier daughters

i. Daughters may be affected if mother is also carrier or affected10. Barr bodies are found in female cells, as one X-chromosome is inactivated

a. One of the two X chromosomes in female cells is in the heterochromatin stateb. Usually females cells will only have one Barr body, but the number can differ due to

aneuploidy11. Autosomal genes can be sex-influenced, with the heterozygotes having different phenotypes

based on gender12. Autosomal genes can also be sex-limited, where only one gender gets affected13. Having the genotype does not necessarily mean the individual is affected

a. Penetrance and expressivity measures thisb. Pleiotropy also leads to not all effects being caused

14. Multiple Alleles will cause an increase in the number of genotypes

a. Number of genotypes = n(n+1)2

, where n is the number of alleles

15. Epistasis occurs when the genotype at one locus suppresses the expression of the alleles of another locus

16. Recessive epistasis requires for a homozygous recessive genotype to mask the other genea. Loci for gene A has recessive epistasis over loci for gene Bb. Caused by the loss of production of intermediate, masking the effect of the other

genei. Like ability for coat to have colour

c. Caused by leaky allele, where two recessive alleles gives rise to an effect similar to a dominant allele

17. Dominant epistasis requires only one copy of the allele at the epistatic locia. The expression of the dominant allele masks the expression of the other gene

i. The dominant allele causes the phenotype regardless of the allelic conditions of the other locus

b. Cause by the dominant allele degrading the precursor or product of the other gene18. Complementary loci (Duplicate recessive gene) is when the recessive alleles for both genes

give the same phenotypea. Like colourness and purple vs white

b. Caused by the pathway having a colourless precursor and colourless intermediatec. Caused by needing both products to form a functional dimer

19. Duplicate dominant gene is when the dominant alleles for both genes give the same phenotype

a. Both genes lead to the same final product, and phenotype is independent on the relative quantity of final produce produced

20. Expression of the gene can be regulated by epigenetic factorsa. Positive or negative control of inducers and repressors for translationb. DNA topoisomerase can affect packing of DNA, affecting access to genec. DNA and chromatin modification, like methylationd. RNA processing can affect stability and export

i. Alternative gene splicing produces different proteinse. microRNA can silence the mRNAf. Protein modificationg. Prionsh. Factors affecting enzyme activity

21. Lethal genes cause the death of the organisma. Dominant lethal genes only allow homozygous recessive organisms to survive

i. The heterozygote might not die immediately at birth, but only develop lethal symptoms later in life

b. Recessive X-chromosome lethal genes will only affect malesi. Because an affected daughter cannot be born

There is no affected father to pass on the disease alleleii. Carrier daughters can be born from their carrier mothers, thus making their

sons vulnerable22. Nondisjunction can occur during either Pachytene of prophase I or anaphase I or II

a. Either the Holliday junction is not resolved leading to a persistent chiasma, or the spindle fibres fail to achieve proper attachment.

b. Variation in chromosome number is called aneuploidyi. Monosomy is one chromosome missing

ii. Trisomy is one chromosome extra23. Uniparental disomy occurs when only one parent contributes both the allele of the gene

a. Cause by disjunction occurring in both parents, where one gives both, and the other gives none

b. Or caused by trisomic rescue, where the cell randomly chooses to eliminate the extra chromosome, removing the non-nondisjuction allele with it.

Non-Mendelian Genetics

1. Maternal inheritance is non-Mendelian because the mitochondrial DNA can only be passed down from mother to offsprings

a. Mitochondrial DNA undergoes high rates of mutation i. Due to exposure to free radicals from respiration

ii. Due to no DNA repair during replicationiii. Due to its numerous copies in each mitochondria

b. Thus can be used for comparison between species that diverged only a short time ago

2. Linked genes are non-Mendelian because they do not obey Mendel’s Law of Segregationa. The genes tend to be inherited together, affecting the offspring genotypic ratio

i. Parental genotype will be greater than recombinant genotypeii. Genes can be on the same chromosome but still be genetically unlinked, if

they are very far apartb. Each chromosome is thus a linkage group, where genes are inherited together

i. Thus humans have 23 linkage groups3. The distance between two genes loci on homologous chromosomes are measured in

centiMorgana. 1 map unit or cM is equal to a 1% chance that random chiasma formation will occur

between the two gene loci on non-sister chromatidsi. Genes with greater cM have a chance of multiple chiamata forming between

themb. 1 cM is roughly equal to 1 million base pairsc. Genes that are more strongly linked have a lower cMd. Maximum value for cM is not 50

i. Although recombination frequency maximum is 50%, it tends to underestimate the actual distance, thus 50cM would give rise to less than 50% recombinant frequency

4. Genetic distance is calculated from the proportion of recombinant offsprings

a. Recombination fraction (θ) = Number of recombinant progeny

Totalnumber of progenyi. 0≤θ≤0.5

ii. 1 cM ≅ 0.01 (θ value) They are not equal when the real distance between loci is great There will be less recombinant frequency than expected, as even

numbers of chiasmata formation between the loci produces a parental gamete

iii. Since the Mendelian ratio is 1:1:1:1, which means 50% parent and 50% recombinant

b. A value of 0.1 would imply that crossing over to form recombinant gametes occurs once every 10 times

c. A value of 0.5 would imply that recombinant gametes are form half the time meaning that the gene are either physically or genetically unlinked

5. For linked genes, a heterozygote (AaBb) can have different forms of the same genotypea. Depends on the parents genotype

i. AABB x aabbii. AAbb x aaBB

b. Trans arrangement is when the mutant alleles are carried by both chromosomesi. Ab + aB

c. Cis arrangement is when one chromosome carries the wild-type allele, and the other carries the mutant allele

i. AB + ab6. Linkage analysis can be done to associate a certain form of a marker with the disease allele

of a genea. Closely linked markers with the disease allele will cosegregate b. Markers can be repetitive DNA sequences with the flanking DNA sequence being

uniquei. Microsatellites are repeats of 2 to 5 nucleotides

ii. Minisatellites are tandem repeats of 15 to 35 nucleotide sequences

Population Genetics

1. A population has a large group of sexually-reproducing interbreeding individuals with a common gene pool

2. Hardy-Weinberg equilibrium is achieved when a population has no changes in genotype frequency, at a specific loci, over generations.

a. Actual numbers of individuals with the genotype may change, but the overall genotype frequency remains constant

b. When a population is in Hardy-Weinberg equilibrium, p2+2 pq+q2=1i. Where p is the frequency of one allele, and q is the frequency of the other

allele p+q=1

ii. The equation means, that at a specific allele frequency, there must be a specific number of organisms that are homozygous dominant ( p2), heterozygous (2 pq ¿, and homozygous recessive (q2)

iii. The equation is a symptom of Hardy-Weinberg equilibrium, not a test, as the population can coincidently be at the equilibrium ratio, but is still in the process of changing.

iv. If the gene has more than 2 alleles, the equation changes p+q+r=1 p2+q2+r2+2 pq+2 pr+2qr=1

c. The values of the genotype frequency, when there is Hardy-Weinberg equilibrium, is not fixed

i. When conditions change from ABC to ABD, and back to ABC, the equilibrium genotype frequency at the first ABC is not the same as during the second ABC

3. Hardy-Weinberg equilibrium is obtained when certain conditions are satisfieda. There must be a large population size, to minimise the effects of genetics drift

i. Since mating and survival is random, the genotype of a small population will fluctuate greatly

Fixation occurs when one allele is lost from the gene pool This is more likely to occur in small populations

b. There needs to be random matingi. Non-random mating leads to increases in the frequency of homozygotes in

the sub-population Like population stratification by religion or ethnicity Like assortative mating where those with desired traits are more

popular Positive assortment, when like only mates with like,

increases frequency of homozygous genotypesHomozygous individuals, while more likely to

expresses desirable traits, are also more likely to expresses undesirable traits

Negative assortment, when like strictly does not mate with like, increases frequency of heterozygous genotypes

Like inbreeding

c. There needs to be no net migration of genotypesi. Movement of individuals can increase or decrease gene frequencies

ii. Or even introduce new genes through gene flowd. There must also be no net mutations

i. The spontaneous mutations of an allele into the other form for an offspring, needs to be counterbalanced by the death of an organism carrying the alternative allele.

ii. No new allele should be formed also, else it will definitely change genotype frequency when passed to an offspring.

e. Lastly, there needs to be no net selection pressurei. Favouring the successful reproduction chances of a genotype will lead to an

increase in the frequency of that genotype over time. And vice-versa

ii. Selections against a genotype can be offset by new mutations.4. Hardy-Weinberg equilibrium is obtained after 1 generation if all assumptions hold true.

a. If all conditions are true for the current parent generation, their offspring will have the exact same genotypic frequency as them, showing that there is Hardy-Weinberg equilibrium

b. This equilibrium is maintained so long as the environment does not change in a way to affect the targeted locus.

5. For sex-linked loci, the allele frequency in males will be equal to the allele frequency in females of the previous generation

a. Males will chase after females, because they are hemizygous and inherit the X-chromosome from the females

6. Mutations will change the alleles of genes, introducing genetic variationa. Non-recurrent mutations are rare events, and not important for population genetics

i. The probability of the rare allele successfully propagating through the population over generations is unlikely

b. Recurrent mutations will introduce new alleles into the population, and sustain them

i. New allele is sustained by a changeable mutation rate from wild type to mutant

If the mutation is one way and recurrent, eventually all the wild type allele will be converted to the mutated form.

The allele frequency of the wild type after n generations: pn=p0 (1−μ )n

Where μ is the mutation rate This leads to an eventual loss of genetic variation

If the mutation is two way and recurrent, a balance between the two forms will be reached

At equilibrium, there will be no net change in gene frequencies, υ qe=μ pe

Where υ is the rate of mutation of allele A1, whose frequency is q, and μ is the rate of mutation of allele A2, whose frequency is p

If given the mutations rates, to find the allele frequency at

equilibrium: qe=μ

υ+μ, pe=

υυ+μ

Note that the numerator is the mutation rate of the other allele

This maintains the genetic variation in a population7. Natural selection will select for fitter organisms to survive to reproduce

a. The fitness of a genotype is 1−si. Where s is the coefficient of selection (selection intensity), the

proportionate reduction in gametic contribution of the genotypeii. The fittest genotype has a fitness value of 1 (s=0)

iii. A fitness of 0.5 means that for every one offspring produced by the less fit genotype, 2 offspring are produced by the fittest genotype

Thus the population has more of the fittest genotypeb. To find the allele frequency when it is selected against

i. Assume that the population is in Hardy-Weinberg equilibrium, such that the observed number are equal to the expected numbers

ii. Calculate the current allele frequency of all allelesiii. Calculate the gametic contribution of each genotype

Genotype frequency times fitnessiv. Derive the allele frequency over one generation

q1=Gametic contributionby homozygous+Gametic contributionof allele by heterozygous

Total gametic contributionby all genotypes Note that heterozygous will only contribute the allele half

the time.v. Derive a general formula for allele frequency

c. If there is overdominance, where the heterozygotes has the greatest fitness, the population can obtain equilibrium

i. s1 pe=s2qe Because at this level of heterozygotes, the number of homozygous

organisms produced is equal to the number that dies, thus maintain the frequency

ii. pe=s2

s1+s2, qe=

s1s1+s2

8. When analysing a population to check for Hardy-Weinberg equilibriuma. Calculate the frequency of the first allele (A1)

i.

p=Number of organisms with A1 A1+

12(Number of organismswith A1 A2)

Total number of organismsb. Calculate the frequency of the second allele (A2)

i.

q=Number of organismswith A2 A2+

12(Number of organisms with A1 A2)

Total numberof organisms

c. Calculate the expected frequency of each genotype if the population is in Hardy-Weinberg equilibrium

i. A1 A1=p2 , A1 A2=2 pq , A2 A2=q2

d. See if the number of organisms corresponds to the expected numbers should the population be in Hardy-Weinberg equilibrium

i. Expected numbers for genotype A1 A1 is ( p2×total numberof organisms)ii. Repeat for the other genotypes

iii. Perform χ2-test

χ2=∑(observed numbers−expected numbers )2

Expected numbersiv. Compare χ2 value with table

Degree of freedom is number of genotype−number of alleles This is because the allele frequencies are not independent of each

other, and fixing the frequency of one determines the other Since p+q=1

9. When deriving an unknown genotype frequency in a population, from a given genotype frequency

a. First assume that the population is in Hardy-Weinberg equilibrium, and that the observed numbers equal the number expected of such a population in Hardy-Weinberg equilibrium

b. Use the equation p2+2 pq+q2=1 to work out the frequencies of the other two genotypes or other allele

c. If the gene is sex-linkedi. Assume that there are equal numbers of male and females in the population

ii. Frequency of known allele in entire population = 23

( Allele frequency∈females )+ 13(Allele frequency∈males)

A population in equilibrium should have allele frequency in entire population = allele frequency in females = allele frequency in males

10. Z-score is used to compare allele frequency between population to see if it significantly differs

a. Z-score =

p1−p2

√ p1 (1−p1 )n1∗2

+p2 (1−p2 )n2∗2

i. Z-score is significantly different if the value is greater than 1.9611. Polymorphism occurs when a population has two or more alleles that are maintained in high

enough frequencies (greater than 1% of population) that the rarest of them cannot be maintained by just constant mutation and selection

a. Other factors play a role in keeping the rarer allele in the gene pooli. Heterozygote advantage

There will be an equilibrium geneotype frequency which is at intermediate levels (above 1%)

ii. Frequency dependent selection Selection intensity for a genotype constantly changes, thus

maintaining a genotype frequency range over time, keeping the rare alleles around

iii. Heterogeneous selection The environment of the population changes, thus changing selection

and mutation rates The environment change is either over time, or by

population movementiv. Transition

Explains some currently observed polymorphisms Due to recent environmental changes, the selection and mutation

values have just changed, and the polymorphism will be lost over time.

v. Neutral mutations The polymorphism confers no selection advantage Thus polymorphism is maintained by random mutation and random

death Polymorphic by chance alone

Greater chance when population size is larger12. The founder effect is when genetic drift occurs through a new population branching off the

original population, bringing with them a subset of alleles of the original populationa. The new population has different genotype frequencies depending on who left the

original population13. A bottleneck event is when the variation in a population’s gene pool is greatly reduced due

to environmental events like natural disasters or diseasesa. Like the Black deathb. Alleles can be lost, and the genotype frequency changes

Questions

1. Can the prophase of mitosis be divided into Leptotene and Diaknesis as well?2. As the microtubules radiate from the centrosome, are they all considered astral until they

contact the kinetochore (making them kinetochore MT) or microtubules from the opposing pole (making them polar MT)

3. Will secondary nondisjunction always occur for those with trisomy?

Resources

1. Chromosome and Kinetochorea. https://www.youtube.com/watch?v=0JpOJ4F4984

2.