pro and eu

Question 1

Comparison between eukaryotes and prokaryotes
cell size
nucleus
genetic material
ribosomes for protein synthesis
organelles
cell wall

Answer 1

larger vs smaller
nucleus with nuclear envelope present vs no true nucleus
linear DNA associated with histone proteins found in nucleus and no plasmids vs circular DNA associated with few histone-like proteins, found in nucleoid region with plasmids present
80s ribosomes vs 70S (no ER)
many membrane bound organelles present vs none
compose of cellulose in plants and chitin in fungi vs peptidoglycan

Question 2

comparison of structure and organisation of genome between pro and eu
size
appearance
molecule
association with proteins
level of DNA packing/coiling
location
extrachromosomal DNA
number of genes
non coding regions
presence of operons

Answer 2

larger vs smaller
multiple linear molecules vs single circular molecule
both are double helix DNA
both are yes but eukaryotes have more histones/histone like proteins
high vs relatively low
nucleus vs nucleoid region
yes if u consider mitochondrial and chloroplast circular DNA + yes for plasmids
25k vs 4.5k
common vs uncommon
few vs many

Question 3

how is DNA coiled in eukaryotes

Answer 3

DNA double helix –> negatively charged DNA is wound around 8 positively charged histone proteins twice to form nucleosomes with linker DNA joining adjacent nucleosomes, forming a 10nm fibre which coils around itself to form 30nm solenoid which forms looped domains when associated with scaffold proteins, forming 300nm fibre which supercoils to form metaphase chromosome

Question 4

how is DNA coiled in prokaryotes

Answer 4

DNA double helix –> folded into looped domains by protein-DNA associations which undergoes supercoiling with the help of DNA gyrase and topoisomerase

Question 5

how does telomerase work

Answer 5

1. A short 3-nucleotide segment of RNA within telomerase binds to part of a DNA repeat in the 3’ overhang by CBP
2. the adjacent part of the RNA within telomerase is sued as a template to synthesise a short complementary 6-nucleotide DNA repeat
3. telomerase catalyses the formation of the phosphodiester bonds between the existing 3’OH group of existing DNA overhang and 5’ phosphate group of incoming deoxyribonucleotide (dNTP)
4. after the repeat is made, telomerase translocates 6 nucleotides to the right in the 5’ to 3’ direction of the DNA overhang and begins to make another repeat. the process is repeated such that a series of tandem repeats are made, elongating the telomere
5. primase makes a RNA primer near the end of the telomere. DNA polymerase adds nucleotides to the 3’OH end of the primer and hence synthesizes a complementary strand. the nick is then sealed by ligase. the RNA primer is eventually removed.

Question 6

what are the non-coding regulatory DNA sequences

Answer 6

introns promoters enhancers silencers

Question 7

what are the non coding repetitive DNA sequences

Answer 7

telomere and centromere

Question 8

structure and function of intron

Answer 8

non-coding DNA sequences found within a gene, specifically between exons in a specific segment of DNA and pre mRNA
enables alternative RNA splicing to occur where a single pre-mRNA can have all its introns and different combinations of its exons excised and the remaining exons joined such that different mature mRNAs are produced.
a gene can code for more than 1 polypeptide

Question 9

structure and function of promoter

Answer 9

located just upstream of the transcription start site of a gene hence called a proximal control element
has critical elements such as TATA box and CAAT and GC boxes
for recognition and binding site for GTFs which then recruit RNA polymerase to form TIC which initiates transcription
TATA box in promoter is where GTFs bind and hence determines precise location of transcription start site
CAAT and GC box improve efficiency of promoter

Question 10

structure and function of enhancer

Answer 10

located far away from the promoter, hence called a distal control element
recognition and binding site for activators (STFs)
increase the frequency of transcription by promoting the assembly of the TIC

Question 11

structure and function of silencer

Answer 11

located far away from the promoter, hence called a distal control element
recognition and binding site for activators (STFs)
increase the frequency of transcription by preventing the assembly of the TIC

Question 12

structure of telomere

Answer 12

found at both ends of linear, eukaryotic chromosomes.
non-coding DNA made up of a sreies of tandem repeat sequences, 5’TTAGGG3’
have a single stranded region at their 3’ ends known as the 3’ overhang

Question 13

function of telomere

Answer 13

telomeres ensure that genes are not eroded and vital genetic information is not lost with each round of DNA replication due to the end replication problem
the last RNA primer on the lagging strand with DNA cannot be replaced with DNA. hence the DNA molecule shortens with each round of replication. telomeres that are non-coding sequences at the ends of linear chromosomes will be lost b4 any vital genetic info is.

they protect and stabilise the terminal ends of chromosomes by forming a loop using the 3’ overhang
this prevents fusing of 2 chromosomes and also prevents the cell’s DNA repair machinery from detecting the chromosome as damaged DNA and trigger apoptosis

telomeres allow their own extension as they have a 3’ overhang which provides an attachment point for the correct positioning of the enzyme telomerase

Question 14

structure and function of centromere

Answer 14

constricted region on chromosome where kinetochore microtubules attach during nuclear division + non-coding DNA made up of a series of tandem repeat sequences
allow for sister chromatids to adhere to each other
allow kinetochore proteins to attach and which in turn allow spindle fibres to attach so that sister chromatids can align along the metaphase plate and subsequently be separated to opposite poles
allows for proper alignment and segregation of chromosomes.

Question 15

what is the purpose of regulation of gene expression

Answer 15

cellular differentiation ; adaptation to changes; conservation of resources and increase in variation of proteome despite limited genome size

Question 16

outline histone deacetylation/acetylation

Answer 16

addition of acetyl group to lysine residues in the histone acetyltransferase removes positive charge on histones and decreases electrostatic attraction between negatively charged DNA and histones, promoter region is more accessible to RNA polymerase and GTFs –> promotes transcription by promoting assembly of tic, and vice versa

Question 17

outline the chromatin remodeling complex

Answer 17

CRC alter the structure of nucleosomes temporarily, can cause DNA to be more tightly coiled around histones preventing access of RNA polymerase and GTF to promoter –> inhibiting Transcription by preventing assembly of TIC and vice versa

Question 18

outline process of DNA methylation

Answer 18

addition of a methyl group by DNA methylases to selected cytosine residues in CG sequences, prevents transcription by blocking the binding of transcription factors at the promoter and hence preventing the formation of the TIC + recruiting DNA-binding proteins (e.g transcriptional repressors, histone deacetylases etc etc

Question 19

How do regulatory proteins regulate transcription

Answer 19

STFs vs activators and repressors

Question 20

outline activation (transcriptional)

Answer 20

activators bind to enhancers and promote assembly of TIC as bending of spacer DNA allows interaction of activators with RNA polymerase and GTFs at the promoter; transcription frequency increases.

Question 21

outline repression (transcriptional)

Answer 21

bind to silencers to prevent assembly of TIC as bending of spacer DNA allows repressors to bind to GTFS hence preventing activators from binding to gtfs, transcription frequency decreases

Question 22

what are the different post transcriptional gene regulation methods?

Answer 22

addition of 5’cap
intron splicing
addition of polyA tail

Question 23

outline the addition of the 5’ cap

Answer 23

addition of a 7-methylguanosine nucleotide is added the to 5’ end of the pre-mRNA, occurs shortly after transcription begins
it helps the cell to recognise mRNA
acts as a signal to export mRNA out of nucleus
stabilises and protects the growing pre-mRNA chain from degradation by ribonucleases

Question 24

outline intron splicing

Answer 24

when the introns are excised and exons are joined together by spliceosomes which recognise the sequences at intron-exon boundaries so that functional proteins can be produced
alternative splicing -> diff exons of a single pre-mRNA can be joined tgt such that different mature mRNAs and hence diff proteins can be produced

Question 25

outline addition of poly A tail

Answer 25

the 3’ end of a pre-mRNA is cleaved by an endonuclease downstream of the polyadenylation signal (AAUAAA). then, a poly-A polymerase will then add a long sequence of adenine nucleotides to 3’ end of the pre-mRNA, forming a poly A tail
it occurs imediately after transcription
acts as a signal to export mature mRNA from degradation by ribonucleases
interacts with eukaryotic initiation factors and the 5’cap for initiation of translation

Question 26

what are the different translational gene regulation methods?

Answer 26

mRNA half life
formation of TIC

Question 27

outline mRNA half-life

Answer 27

mRNA half life is determined by the length of its poly-A tail. the longer the poly-A tail, the longer the time mRNA can be used as a template to make proteins. the poly-A tail is removed by ribonucleases in the 3’ to 5; direction until a critical length is reached which will trigger removal of the 5’ cap and degradation of the mRNA from the 5’ end

Question 28

outline the formation of the transcription initiation complex

Answer 28

1: antisense RNAs which are complementary to part of the mRNA to be degraded will bind to the mRNA. the DS RNA will block translation of the mRNA and be targeted for degradation by nucleases
2: during translation initiation, small ribosomal subunit, eukaryotic initiation factors and initiator tRNA form a complex which binds to the 4’cap and teh 3’ poly A tail causing the mRNA to circularise. this can be prevented by binding of translational repressor to 5’cap, 5’ UTR or 3’ UTR and interferes with translation initiation
3: during translation initiation, eukaryotic TIF facilitate the binding of the small ribosomal subunit to the 5’ cap. the availability of activated eukaryotic translation initiation factors is determined by whether or not they are phosphorylated.

Question 29

what are the different post translational gene regulation methods?

Answer 29

Formation of functional proteins
regulation of protein activity
protein degradation

Question 30

outline the Formation of functional proteins
regulation of protein activity
protein degradation

Answer 30

covalent modification of polypeptides make them functional proteins
phosphorylation of eukaryotic TIFs can activate the protein and hence upregulate its activity and vice versa
protein degradation by proteasomes determines how long a protein remains in a cell to carry out its function. tagged with ubiquitin by ubiquitin ligase to be recognised by proteasomes. the proteasomes degrade the proteins into peptides using enzymes.

Question 31

what are the causative factors for cancer

Answer 31

environmental factors –> exposure to carcinogens and ionizing radiation can cause mutations that can lead to cancer
loss of immunnity due to infection with certain viruses -> HIV can weaken the immune system and reduce teh body’s ability to fight infections by otehr virsus
genetic predisposition –> due to Gene mutations which we inherit from our parents
Age–> accumulation of mutations in a cell over a lifetime

Question 32

what are proto-oncogenes and what is the known example

Answer 32

code for proteins that stimulate normal cell division
mutated oncogenes increase the amt of proto-oncogenes protein product by a pt mutation in base sequences of regulatory elements -> lead to uncontrolled cell division
gene amplification, where the number of cpies of a proto-oncogene is increased due to a mistake made during DNA replication . this can lead to excessive production of POG pdt and can lead to uncontrolled cell division
chromosomal translocation such that the proto-oncogene ends up under the control of an enhancer
or by retroviral integration where silencer is inactivated or viral enhancer is inserted or a viral homologue of POG is inserted
increase the intrinsic activity of the POG protein pdt by a pt mutation within the POG; changing the aa sequence of teh POG protein which can become hyperactive or more resistant to degradadtion
e.g ras gene

Question 33

what are tumour suppressor genes and what is the known example

Answer 33

genes that code for protein products that inhibit cell division and rpevent uncontrolled cell division by upregulating the expression of genes involved in cell cycle arrest, DNA repair and apoptosis. mutations could be in promoter or coding sequence of the gene (both copies need to be mutated)
e.g p53 codes for a STF that can activate genes involved in cell cycle arrest; DNA repair and initiation of apoptosis when DNA dmg is beyond repair

Question 34

what are gain in function and loss in function mutations

Answer 34

mutation in js one copy of the allele results in …. thus said to be dominant
mutations in both copies of the allele necessary for the …. thus recessive

Question 35

why is development of cancer a multi-step process
the whole shabangggg

Answer 35

the development of cancer requires the accumulation of mutations in the genes which control regulatory checkpoints of teh cell cycle in a single cell
this will disrupt the normal cell cycle, thus causing the cell to undergo excessive cell proliferation
a gain in function is a dominant mutation where mutation in just one allele of a proto-oncogene will result in its overexpression which will result in the production of excessive amounts of growth factors or hyperactive growth factors leading to excessive cell proliferation
loss of function mutation is a recessive mutation where mutations in both alleles of a TSG will result in the non-functional protein which will disrupt their ability to inhibit cell cycle, enable DNA repair and promote apoptosis
activation of the genes coding for telomerase result in telomeres being lengthened allowing the cell to divide indefinitely
loss of contact inhibition will enable the cells to grow into a benign tumour
angiogenesis occurs within the tumour so that blood vessels formed can transport oxygen and nutrients for its growth.
presence of blood vessels can result in the formation of a malignant tumour capable of metastasizing to other parts of the body via the bloodstream to form secondary tumour.

Question 36

differences between normal and cancer cells
nuclei
genes
cells
cell division
apoptosis
contact inhibition
differentiation
metastasis
angiogenesis

Answer 36

abnormal nuclei with high nucleo-cytoplasmic ratio
oncogenes and mutated TSG present
vary in shape and sie
undergo excessive uncontrolled cell division and proliferation
do not show contact inhibition –> multiple layers of cells
do no differentiate
can detach from surrounding cells and metastasize
stimulate growth of new blood vessels