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DNA Flashcards

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Biology 101 flashcards
1
Q

what are the molecules that have CBP with 2 H bonds

A

ATU

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2
Q

what are the molecules that have CBP with 3 H bonds

A

GC

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3
Q

what are purines and pyrimidines

A

AG and CTU

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4
Q

Define a gene mutation

A

`An alteration in the sequence of nucleotides which may change the sequence of Aas in a ppt chain. this may change the 3D shape and hence the function of the protein and thus the phenotype of the organism; thus mutations can result in new alleles

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4
Q

whats the structure of a nucleotide

A

a nitrogenous base attached to C1
a phosphate group attached to C5
and a pentose sugar

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5
Q

structure of RNA and DNA

A

DNA: constant width between sugar phosphate backbone = 2nm
strands run antiparallel; one complete turn of the double helix has 10BP and a distance of 3.4nm
RNA–> no fixed AUCG ratio; ss

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6
Q

what is semi-conservative DNA replication

A

Both strands of DNA separate by the breaking of hydrogen bonds and each strand acts as a template for the synthesis of a new strand through CBP

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7
Q

types of mutation

A

substitution inversion insertion deletion

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8
Q

what is an inversion mutation

A

segment inverts

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8
Q

what is a substitution mutation

A

replacement of a nucleotide by another

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9
Q

what is a deletion mutation

A

nucleotides are removed from a sequence

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9
Q

what is a insertion mutation

A

nucleotides are inserted into the sequence

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10
Q

whats a frame-shift mutation

A

it is due to an insertion or deletion of a number of nucleotides that is indivisible by3, disrupting the reading frame and resulting in a completely different and non-functional ppt chain formed.

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11
Q

what is a miss-sense mutation

A

point mutation that results in a codon that codes for a diff aa; if similar then no change and vv.

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12
Q

what is a silent mutation

A

a point mutation that does no change the aa sequence in a ppt as the genetic code is degenerate

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13
Q

what is a nonsense mutation

A

a point mutation that results in a premature stop codon UAG UAA UGA causing the ppt to be truncated and non-functional

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14
Q

outline how sickle cell anemia arises due to a mutation

A

affects beta globin chain of haemoglobin
due to a substitution mutation CTC to CAC; the change in mRNA GAG to GUG; there is a change in amino acid from glutamate to valine;
the charged and hydrophilic glutamate is replaced by non-polar hydrophobic valine in HBS; at low oxygen concentrations, HbS will lose the oxygen and undergo a conformation change which will cause the hydrophobic patches on HbS to stick out.
the hydrophobic areas of the different HbS will stick tgt and the polymerisation of HbS results in the formation of abnormal, rigid, rod-like fibres which will distort the shape of the biconcave RBC and make it sickle shaped. when oxygen binds to HbS it is reversed.

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15
Q

wtf are chromosomal aberrations

A

variation in chromosomal structure or number
same as mutations; but deletions and duplications can result in phenotypic abnormalities due to the reduced ro additional genes respectively. inversions and reciprocal translocations can result in disease, as although the amount of genetic material remains the same, gene expression is influenced.

16
Q

what is non disjunction

A

homologous chromosomes do not move properly to opp poles during meiosis1 or the sister chromatids fail to separate properly to opp poles during meiosis 2

17
Q

effects of sickle cell anemia

A

RBC are more fragile and break easily, resulting in a shortage of RBCs and poor oxygen transport. this leads to anemia, lack of energy and heart failure. they may also lodge in small blood vessels and interfere with blood circulation leading to organ damage

18
Q

differences between replication transcription and translation

A

location
where it begins at and ends at
template
monomers
CBP occurs between?
enzymes involved
bonds formed
ribosomes
template strand is read in
molecule is synthesized in ?
proofreading
products
product destination

19
Q

whats the role of DNA

A

store information and pass it on from one generation to the next
this is because it can be replicated accurately (CBP allows them to act as a template)
it is a stable molecule (PPD and HBs)
backup of code due to template strand
coded information can be readily utilised (weak HB allows for easy separate)

20
Q

whats the role of mRNA

A

serves as a messenger that takes information out of the nucleus via the nuclear pore to the cytoplasm where translation takes place
it acts a as template for translation
the sequence of codons will determine the ppt sequence

21
Q

whats the role of tRNA

A

they bring in specific amino acids in a sequence corresponding to the sequence of codon in mRNA to the growing ppt.
it facilitates translation due to the ability to bind to a specific single aa during aa activation
and the ability of the anticodon to base-pair with the mRNA codon during translation

22
Q

whats the role of rRNA

A

1) rRNA associates with a set of proteins to form ribosomes
2)rRNA is the main constituent of the interface between the large and small subunits of the ribosome thus the small ribosomal subunit can bind to the mRNA as CBP can occur between the rRNA in the mRNA binding site of the small ribosomal subunit and the mRNA
3) rRNA is the main constituent of the P site and A site on the large ribosomal subunit hence it enables the binding of aminoacyl-tRNAs to the P and A site by CBP
4) peptidyl transferase on the large ribosomal subunit also catalyses the formation of the peptide bond between the amino group of the new aa in the A site and the carboxyl end of the growing ppt in the P site

23
Q

features of the genetic code

A

it is a triplet code
universal ; degenerate; non-overlapping; continuous ; includes start and stop codons (AUG UGA UAA UAG)

24
Q

what are the main points of the pathway of DNA replication

A

Unzipping of parental strand
addition of primer
synthesis of daughter strands
end of replication

25
Q

what happens before unzipping of parental strand

A
  1. before DNA replication, free dNTPs are manufactured in the cytoplasm and transported into the nucleoplasm via nuclear pores.
  2. DNA replication occurs at S phase of interphase
25
Q

outline the unzipping of the parental strand

A
  1. replication begins at specific points of the DNA molecule each of which is known as an origin of replication.
  2. helicase binds to origin of replication. it disrupts hydrogen bonds between CBPs, causing parental strands to unzip and separate. this process requires ATP.
  3. single-strand binding proteins keep the strands apart so that they can serve as templates for the synthesis of new strands
  4. topoisomerase relievers ‘overwinding’ strain ahead of replication forks by breaking, swiveling and rejoining DNA strands.
26
Q

outline the addition of primer

A
  1. RNA primer is added to each template strand by the enzyme primase
  2. RNA primer provides a free 3’ OH end for DNA polymerase to recognise and start DNA synthesis of the Complementary daughter strand.
  3. DNA polymerase can only add deoxyribonucleotides to a pre-existing 3’OH end of a nucleotide
27
Q

outline the synthesis of daughter strands

A
  1. DNA polymerase uses the parental strand as a template an aligns the free activated dNTPs in a sequence complementary to that of the parental strand.
  2. A pairs with T and G pairs with C
  3. DNA polymerase catalyses the formation of ppd bonds between adjacent daughter DNA nucleotides of the newly synthesised strand
  4. As DNA polymerase moves along the template, it proof reads the previous region for proper BP. any incorrect deoxyribonucleotide is removed and replaced by the correct one
  5. The leading strand is synthesised continuously in the 5’ to 3’ direction
  6. the lagging strand is synthesized discontinuously. the lagging strand is synthesised in fragments known as okazaki fragments
  7. a different DNA polymerase then removes the RNA primer and replaces it with deoxyribonucleotides.
  8. DNA ligase seals the nicks by forming PPD between adjacent nucleotides of each of the DNA fragments on the new strand.
28
Q

outline the end of replication

A
  1. complementary parental and daughter strands rewind into a double helix
  2. each resultant helix consists of one parental strand and one daughter strand hence this is called semi-conservative DNA replication.
29
Q

termination points
transcription

A

after RNA polymerase transcribes through the termination sequence, the mRNA chain is released and the RNA polymerase will dissociate, terminating transcription

29
Q

initiation points
transcription

A
  1. RNA polymerase attach to the promoter with the aid of protein factors
  2. RNA polymerase unzips and separates the DNA double helix at promoter by breaking HBs between CBPs.
  3. only one strand is used as the template to synthesize complementary mRNA strand
29
Q

whats the main points of transcription pathway

A

synthesis of RNA using a DNA template
Initiation
elongation
termination
post-transcriptional modification
(occurs in nucleus)

29
Q

elongation points
transcription

A
  1. Free ribonucleotides will bind by CBP to deoxyribonucleotides on DNA template strand
  2. AUGC binds to TACG
  3. Cytosine forms 3 HBs with G, A forms 2 HBs with T and U
  4. RNA polymerase catalyses the formation of covalent ppd bonds between adjacent ribonucleotides, forming the sugar-phosphate backbone of the growing mRNA transcript
  5. mRNA strand is synthesized int he 5’ to 3’ direction as new ribonucleotides are added to the 3’OH end of the growing mRNA strand
  6. As the transcription complex continues to move down the DNA double helix, unzipping the 2 strands, and synthesizing mRNA, the region of DNA that has just been transcribed, reanneals.
30
Q

post transcriptional modification points

A
  1. Addition of 7-methylguanosine cap to 5’ end of pre-mRNA
    protects mRNA from degradation by ribonucleases, facilitates the export of mature mRNA from the nucleus to the cytoplasm and helps the cell recognise mRNA so that it can undergo subsequent steps such as splicing
  2. RNA splicing involves spliceosomes which excise introns and join exons
  3. synthesis of poly A tail involves the cleaving of the pre-mRNA by an endonuclease 10-35 nucleotides downstream of teh polyadenylation signal and the addition of many adenine nucleotides by the enzyme poly A polymerase downstream of the polyadenylation signal, AAUAAA.
    protects mRNA from degradation by ribonucleases, facilitates the export of mature mRNA from the nucleus to the cytoplasm and interacts with initiation factors to form the TIC.
30
Q

what is translation and the main points

A

synthesis of a ppt using genetic information encoded in an mRNA molecule
initiation
elongation
termination

31
Q

How does amino acid activation occur

A

1) a specific amino acid is covalently attached to the 3’ CCA stem of a specific tRNA with a specific anticodon by a specific aminoacyl-tRNA synthetase.
2) there are 20 different aminoacyl tRNA synthetases. Each enzyme has an active site which recognizes a specific aa and the unique identity sites at the 3’CCA stem and the anticodon loop of a specific tRNA

32
Q

Initiation of translation

A

translation initiation factors will facilitate binding of small ribosomal subunit and initiator tRNA carrying methionine to the newly synthesized mRNA strand
1) the anticodon of initiator tRNA with complementary base pair with start codon (AUG) of mRNA
2) binding of large ribosomal subunit will complete ribosome forming TIC
3) this positions initiator tRNA at peptidyl-tRNA binding site (P site) leaving aminoacyl-tRNA binding site (A site) vacant for incoming aminoacyl-tRNA molecules.

33
Q

elongation of translation

A

A) codon recognition: a second aminoacyl-tRNA with a specific anticodon and corresponding amino acid, CBPs with a specific mRNA codon at the A site by forming HBs
B) a peptide bond is formed between adjacent aas, catalysed by peptidyl trasnferase on the large ribosomal subunit
methionine dissociates from initiator tRNA and as a result remains bound to second aa at A site
C) the ribosome translocates in 5’ to 3’ direction, shifting first tRNA to exit site (E site) allowing it to be released into cytosol
tRNA with growing polypeptide chain is now at P site
A site will hold a new incoming aminoacyl-tRNA with an anticodon complementary to next codon on mRNA
process is repeated until a stop codon is reached.

34
Q

eukaryotes vs prokaryote difference

A

in eukaryotes, transcription takes place in the nucleus and the pre-mRNA undergoes post-transcriptional modification within nucleus before being transported to the cytoplasm for translation
in prokaryotes, mRNA can be translated while it is being transcribed.

34
Q

termination of translation

A

1) termination begins when the stop codon reaches the A site
2) release factors will enter A site causing hydrolysis of covalent bond between polypeptide chain and tRNA in P site
3) polypeptide is released from ribosome and will fold into its secondary and tertiary structures
4) the ribosome disassembles into its subunits.

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