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Alevel Biology > DNA & Genomics > Flashcards

DNA & Genomics Flashcards

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1
Q

Sugar-phosphate backbone shows directionality. What is meant by the 5’ end and 3’ end of the chain?

A

5’: free phosphate group attached to C5 of deoxyribose
3’: free hydroxyl group attached to C3 of deoxyribise

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2
Q

Compare the structure of DNA and RNA.

A
  1. Location
  2. Amount
  3. Macro features
    - DS vs SS
    - Chemical stability
    - Size
    - directionality (S)
    - Sugar phosphate backbone (S)
  4. Variation/forms
  5. Nt
    - basic unit
    - sugar residue (S/D)
    - nitrogenous bases
    - ratio of bases
    - bond (S/D)
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3
Q

What are the advantages of DNA being double stranded?

A
  • DNA mlcs more stable in structure as collective H bonds between base pairs strengthen double-helical structure
  • One strand can act as a template for DNA repair of the other strand
  • DNA can be replicated accurately via semi-conservative replication (each strand serves as a template for synthesis of a new complementary strand)
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4
Q

What is the importance of Chargaff’s rule/1:1 ratio in the structure of DNA?

A

⇒ indicate cbp, where A pairs w T, G pairs w C
⇒ H bonds between complementary bases stabilize struc of DNA
⇒ pairing between purine & pyrimidine→ constant width of 2.0nm between 2 sugar-phosphate backbone

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5
Q

What is the role of DNA?

A

Store info, pass it from one generation to the next

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6
Q

What makes DNA a suitable store of information?

A
  1. Can be accurately replicated: daughter cells have identical copies of DNA as parent cell
    - Weak H bonds between 2 strands→ can separate & act as templates for new strand synthesis
    - Complementary base pairing: A=T, G≡C
  2. Stable molecule→ passed on to next gen wout loss of coded info
    - Collectively, numerous H bonds hold 2 DNA strands tgt
    - adj nucleotides in each strand joined by strong covalent phosphodiester bond
  3. Backup of the code
    - DNA is ds
    - One strand can serve as template for the repair of the other, when mutations occur on either one
  4. Coded info readily utilised/accessed
    - Weak H bonds between 2 strands→ template strand can separate from non-template strand allowing transcription to take place
    - Complementary base pairing: faithful transfer of info from DNA to RNA during transcription, which will subsequently be translated to protein
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7
Q

Describe the role of mRNA in protein synthesis.

A
  • synthesized from transcription, by cbp w DNA serving as a template, where A=U, T=A, G w C
  • Acts as a messenger and carrier of genetic code, to carry transcripts of info from gene in nucleus to ribosomes in cytoplasm/RER, where translation takes place, via nuclear pores
  • template for translation
  • each codon in the coding region specifies an aa in a polypeptide chain→ seq of codons on mRNA code for aa seq in a single polypeptide
  • During translation:
    > codons in mRNA cbp with anticodons of a specific tRNA carrying a specific aa
    > mRNA has recognition sites that allow it to bind to the small ribosomal subunit
    > mRNA has start/stop codons
  • gene expression can be regulated (varying the rate of mRNA synthesis/rate of breakdown)
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8
Q

Describe the role of tRNA in protein synthesis.

A

Brings in specific aa (to the growing polypeptide at ribosome) in a seq corresponding to seq of codons in mRNA.
It can facilitate translation because:
- 3’ end w CCA stem→ ability of tRNA w specific anticodon to bind to specific single aa during aa activation
- ability of its specific anticodon to base-pair to a specific mRNA codon during translation (A bp w U; G bp w C, by forming H bonds)→ ensures seq of nt on mRNA translated into a specific seq of aa in polypeptide chain.

tRNA is then released after aa joins polypeptide, and reused, attaching to another specific aa

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9
Q

Describe the role of rRNA in protein synthesis.

A
  1. associates w ribosomal proteins→ ribosomes
  2. small subunit: cbp between rRNA in mRNA binding site and mRNA, where A base pairs w U, G base pairs w C→ small subunit binds to mRNA
  3. large subunit
    - enables binding of aminoacyl-tRNAs to Peptidyl site (P site) & Aminoacyl site (A site) via cbp
    - part of it acts as peptidyl transferase, catalysing formation of peptide bond between 2 aa
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10
Q

What is the function of ribosomes?

A

Synthesise polypeptide under direction of mRNA by:
- hold tRNA and mRNA closely, allow interaction between codon of mRNA & anticodon of tRNA
- positions new aa for addition to growing polypeptide
- peptidyl transferase catalyses formation of peptide bonds between 2 aa

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11
Q

What’s the difference between ribosomes in prokaryotes and eukaryotes?

A
  • eu: 80S (small (40S) + large (60S))
  • pro: 70S (small (30S) + large (50S))
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12
Q

Define gene.

A

a specific seq of nucleotides in a DNA molecule, which codes for a specific seq of aa in one polypeptide chain. It’s located in a fixed position (locus) on chromosome/DNA mlc, and specifies a particular biological function (phenotype).

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13
Q

Template strand/antisense strand = non-coding seq, mRNA complementary to it. True/False?

A

True

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14
Q

Outline the features of the genetic code (+ definition)

A

Describes the manner in which a particular nucleotide sequence is translated into its corresponding amino acid sequence

  1. Triplet code, where each seq of 3 consecutive nt/codon codes for 1 aa
    - If 1 nt coded for 1 aa→ only be 4 aa; If 2 nt code for 1 aa→ only be 16 (42) aa; If 3 nt code for 1 aa→ only be 64 (43) aa, > enough to code for all 20aa
  2. Universal: same triplet of nt codes for same aa in all organisms (basis of genetic engineering)
  3. Degenerate: For some aa out of the 20, an aa may be coded for by >1 codon
  4. Non-overlapping: Codons read as successive groups of 3 nucleotides
  5. Continuous: no nucleotides ‘skipped’ between codons; code read as continuous seq of nt bases
  6. Includes ‘start’ & ‘stop’ seq
    - Start codon: AUG→ signals initiation site for translation of mRNA into seq of aa. It codes for methionine
    - Stop codon: UAA/UAG/UGA→ don’t code for any aa; stop signals for termination of polypeptide chain synthesis during translation
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15
Q

What is conservative DNA replication?

A

2 parental strands separate, act as templates for synthesis of new strands→ reassociate→ restored original double helix + daughter DNA molecule consisting of 2 newly synthesised strands

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16
Q

What is dispersive DNA replication?

A

parental DNA fragmented and dispersed→ daughter molecules w mixture of old & newly synthesised parts

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17
Q

Describe semi-conservative DNA replication.

A

parental DNA molecule separates into 2 single strands through breakage of H bonds, each act as template for synthesis of complementary daughter/new strand through cbp→ 2 new DNA molecule, each a hybrid of 1 parental strand & 1 newly synthesised strand

18
Q

What is the advantage of having multiple Ori?

A

DNA polymerase can add dNTPs at a certain max rate (fixed). Multiple Ori in a single DNA molecule→ many DNA pol can work simultaneously→ speed up copying of very long DNA mlcs

19
Q

Describe DNA replication: START

A
  • replication starts at Ori
  • Helicase breaks H bonds between complementary base pairs of 2 strands→ unzips DNA double helix & separates the 2 parental DNA strands
  • single-stranded binding proteins bind to and stabilise separated ssDNA, keeping them apart so that they remain ss and can serve as templates for synthesis of complementary DNA strand
  • Topoisomerase relieves overwinding strain ahead of replication forks by breaking, swivelling & rejoining DNA strands
20
Q

Describe DNA synthesis: synthesis of new strands

A
  • Primase catalyses synthesis of RNA primer on each parental DNA strand→ provides free 3’ OH needed for DNA polymerase to initiate DNA synthesis
  • Cbp occurs between template strand and free incoming dNTPs, where A forms 2 H bonds w T and G forms 3 H bonds w C
  • DNA polymerase catalyses the formation of phosphodiester bonds, linking DNA nucleotides & synthesizing the new strand in the 5’ to 3’ direction
  • Anti-parallel parental strands→ 2 new strands synthesised in opposite directions
    > Leading strand synthesised continuously towards replication fork
    > Lagging strand synthesised discontinuously away from replication fork, forming Okazaki fragments
  • DNA polymerase (part of it) “proof-reads” previous region as it moves along the parental strand→ ensures proper base pairing between bases. If incorrect DNA nt added, it’s removed by DNA polymerase and replaced w correct one→ ensures fidelity of DNA seq
  • Different DNA pol removes RNA primer and replaces it w DNA nucleotides
  • DNA ligase catalyse formation of phosphodiester bond between nt of adjacent DNA fragments, sealing nicks in DNA
21
Q

Compare DNA replication in eukaryotes and prokaryotes.

A
  1. When
  2. Where
  3. No. of Ori
  4. Ends at
  5. Rate
22
Q

Describe transcription

A

Process: Initiation
- RNA polymerase attach to promoter of gene w aid of transcription factors (protein)→ transcription initiation complex
- RNA polymerase unzips DNA double helix (and separates the 2 strands of DNA) by breaking H bonds between complementary base pairs.
- 3’ to 5’ strand/1 strand used as template strand to synthesise complementary mRNA strand

Process: Elongation
- Free ribonucleotides will bind by cbp to nt on DNA template strand: A=U, T=A, G≡C
- RNA polymerase catalyses formation of phosphodiester bonds between free ribonucleotides→ sugar phosphate backbone
- New mRNA strand synthesised in 5’ to 3’ direction
- Region of DNA that has been transcribed reanneals, forming double helix

Process: Termination
RNA polymerase dissociate from template DNA strand after it transcribes the termination seq
ss pre-mRNA molecules released

23
Q

Describe post-transcriptional modification: capping

A

add 5’ methylguanosine cap→ methylated guanosine nucleotide added to 5’ end of pre-mRNA as soon as ~25 nucleotides produced (protects from degradation by ribonucleases, export of mature mRNA, recognition of mRNA)

24
Q

Describe post-transcriptional modification: splicing

A
  • introns excised, exons joined
  • done by spliceosome with high accuracy
25
Q

Describe post-transcriptional modification: polyadenylation

A

Add poly A tail
- Cleaving of 3’ end of pre-mRNA by enzyme endonuclease 10-35 nt downstream of polyadenylation signal, AAUAAA (highly conserved)
- Poly-A polymerase adds many adenine nt (AMP) to 3’ end to form poly-A tail

26
Q

What happens during amino acid activation?

A

Aminoacyl-tRNA synthetase catalyses the covalent attachment of specific aa to 3’CCA stem of a tRNA with a specific anticodon, to form aminoacyl-tRNA

27
Q

Explain how correct aa is joined to tRNA

A

Aminoacyl-tRNA synthetases have specific AS that are complementary in 3D conformation & charge to specific aa to be attached to tRNA & specific tRNA w specific anticodon→ enzyme achieves double specificity

28
Q

Describe translation

A

Process: Initiation
- Translation initiation factors facilitate binding of small ribosomal subunit and initiator tRNA carrying methionine to newly synthesized mRNA strand. Anticodon of initiator tRNA will cbp with start codon of mRNA.
- Binding of large ribosomal subunit→ tgt forms translation initiation complex
- This positions initiator tRNA at P site, A site vacant for incoming aminoacyl-tRNA
- GTP needed (GTP→ GDP)

Process: Elongation & Translocation
- GTP needed
- 2nd aminoacyl-tRNA w specific anticodon and corresponding aa binds to A site via cbp (GTP→ GDP)
- Peptide bond formed between adjacent aa, methionine & 2nd aa in A site, catalysed by peptidyl transferase in large subunit of ribosome: 1st aa dissociates from initiator tRNA it’s originally bound to
- Translocation (GDP→ GTP): Ribosome translocates one codon down mRNA in 5’ to 3’ direction
> 1st tRNA shifted to E site, released into cytosol and recycled
> 2nd tRNA, carrying growing polypeptide chain, shifted to P site, carries growing polypeptide
> Empty A site: receive 3rd aminoacyl-tRNA, w anticodon complementary to 3rd codon of mRNA
- Process repeats until stop/termination codon on mRNA exposed to A site on ribosome

Process: Termination of translation
- Stop codon reaches A site→ release factors enter A site→ hydrolysis of bond between polypeptide chain & tRNA in P site→ polypeptide released from ribosome & completes its folding into its necessary secondary & tertiary struc (3D conformation)
- Ribosome disassembles into its subunits

29
Q

Why is simultaneous transcription and translation possible in prokaryotes?

A

No membrane bound nucleus (as soon as mRNA is transcribed ribosomes can attach to mRNA for translation) + mRNA no need undergo post-transcriptional modification

30
Q

Compare DNA replication, transcription and translation.

A
  1. Location in cell
  2. Start and end at
  3. Things involved:
    > Enzymes
    > raw material
    > ribosomes
    > Template (S/D)
  4. Process
    > Cbp (S/D)
    > Bonds
    > direction
  5. products
31
Q

Name 2 mutagens

A

excessive UV, tar in cigarettes

32
Q

Explain how gene mutations may affect the protein coded for by the gene

A

change in the nt sequence in DNA→ may change aa seq, which may change 3D shape and hence function of protein→ affect phenotype

Frame shift mutation–> diff aa seq–> diff and non-functional polypeptide–> diff 3D conf of protein; possibly truncated polypeptide

Substitution mutation: one nt replaced by another nt
- Change in codon–> change in aa
- same biochemical properties–> same 3D conf
- diff biochemical properties–> diff 3D conf, may not be functional

Inversion: segment of nt seq separates and rejoins at original position, but inverted
- diff codons on mRNA–> diff aa seq–> diff folding & 3D conf, possibly truncated polipeptide

33
Q

What type of mutation causes sickle-cell anaemia?

A

Single base substitution mutation–> CTC becomes CAC

34
Q

Sickle-cell anaemia: explain the significance of the change in aa to the properties of haemoglobin

A
  • → change in 1°/aa seq in polypeptide: charged, hydrophilic glutamic acid becomes non-polar, hydrophobic valine→ change in bonds between R groups, change in 2°, 3°, 4° structure & change in conformation→ normal Hb A becomes Hb S
  • → change in properties of Hb and phenotype of rbc: At low [O2], O2 released by Hb S, unusual conf change occurs→ hydrophobic patch sticks out, which attaches to another hydrophobic patch on another Hb S→ Hb S polymerises into abnormal, rigid rod-like fibres
35
Q

Sickle-cell anaemia: Describe effects of the change in properties of haemoglobin

A
  • Long insoluble fibres in rbc→ distort shape of normal biconcave rbc and make it sickle shaped
  • Sickle rbc more fragile & break easily→ shorter life span→ shortage of rbc & poor oxygen transport→ anaemia, lack of energy & heart failure
  • Sickle rbc pointed & elongated→ lodge in small blood vessels, interfere w blood circulation
    > Deprive organs of oxygen → organ damage
    > Many localised blockages→ death of surrounding tissue→ severe pain
36
Q

Describe the inheritance of sickle-cell anaemia

A

Recessive condition, both alleles of β globin chains mutated for symptoms to appear
- homozygous recessive: 2 copies of Hb S gene → sickle cell disease
- Heterozygous individuals: Hb A & Hb S→ sickle cell trait; resistant to malaria

37
Q

What kinds of structural chromosomal abberations are there?

A
  • Deletion, duplication, inversion of a segment
  • reciprocal translocation: moves a segment from one chromosome to to another, non-homologous one
38
Q

When is deletion and duplication of a chromosome likely to occur? How does it occur?

A

During crossing over: chromatids break & rejoin at incorrect places→ one chroma give up more genes than it receives→ 1 chromo w deletion mutation and 1 w duplication mutation
→ reduced/additional genes→ phenotypic abnormalities

39
Q

Define aneuploidy

A

cell does not have a chromosome number that is an exact multiple of haploid no.
Extra/fewer copies of chromosomes than wild types
(e.g. trisomic/monosomic)

40
Q

How does non-disjuction lead to aneuploidy?

A

In gamete formation: homologous chromosomes don’t move properly to opp. poles at meiosis I / sister chromatids fail to separate properly to opp. poles at meiosis II
→ 1 gamete receives 2 of the same type chromosome, another gamete receives no copy

If either aberrant gametes unite w a normal gamete at fertilisation, offspring will have abnormal no. of a particular chromosome = aneuploidy. Mitosis subsequently transmits the mutation to all cells of the embryo that arose from this mutant cell

41
Q

What chromosomal abberation occurs in Down syndrome/Trisomy 21?

A

Extra chromosome 21, each body cell w 47 chromosomes
Usually caused by non-disjunction during meiosis I

42
Q

What are the physical characteristics of those with Down syndrome?

A

Characteristic facial features, short stature, heart defects and mental retardation. Most are sterile

Alevel Biology (16 decks)

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