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A level bio > DNA & genomics > Flashcards

DNA & genomics Flashcards

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Biology 101 flashcards
1
Q

Structure of DNA (1-5)

A
  1. DNA molecule consists of 2 polynucleotide chains spiralled around an imaginary axis to form a double helix
  2. The polynucleotide chains are antiparallel
  3. DNA molecules have a uniform width of 2nm
  4. The nitrogenous bases are stacked 0.34nm apart and the helix makes one full turn every 3.4nm, such that there are 10 base pairs in each turn
  5. The 2 hydrophillic sugar-phosphate backbones are on the outside of the helix
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2
Q

Structure of DNA (6-10)

A
  1. The hydrophobic nitrogenous bases are paired in the interior of the helix
  2. The stacking of base pairs results in hydrophobic interactions
  3. The 2 polynucleotide chains are held together by hydrogen bonds between the paired nitrogenous bases
  4. The sequence of bases along one strand is complementary to the sequence of bases along the other strand
    -> adenine pairs with thymine with 2 h bonds via cbp
    -> guanine pairs with cytosine with 3 h bonds via cbp
  5. The 2 grooves between the backbones are called the major groove and minor groove
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3
Q

Function of DNA

A
  1. DNA is the genetic material that organisms inherit from their parents
  2. Most DNA molecules are very long and each DNA molecule contains numerous genes
  3. The gene is a unit of inheritance, which store coded instructions for the synthesis of specific molecules like protein and RNA
  4. During cell division, replication of DNA occurs
  5. During cell division, the 2 strands of DNA separate. Each strand serves as a template from which a new complementary strand is made (semi-conservative replication)
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4
Q

Structure of RNA

A
  1. RNA is a single polynucleotide chain
  2. RNA is less stable than DNA because it is more prone to synthesis by intracellular enzymes
  3. RNA molecule can contain 4 nitrogenous bases similar to those in DNA except that in RNA, thymine is replaced by uracil
  4. RNA can develop secondary structures which are formed by cbp within the RNA molecule. Secondary structures function to promote stability of the molecules
    -> adenine pairs with uracil with 2 h bonds
    -> guanine pairs with cytosine with 3 h bonds
  5. Different forms of RNA include messenger RNA, transfer RNA and ribosomal RNA
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5
Q

Structure & role of mRNA

A
  1. mRNA molecule is a single-stranded RNA
  2. mRNA is synthesised by the transcription of DNA in the nucleus
  3. mRNA is transported to cytoplasm for translation by ribosomes
  4. mRNA is used as a template to synthesise proteins
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6
Q

Structure & role of rRNA

A
  1. rRNA molecule is a single-stranded RNA
  2. rRNA is synthesised in the nucleolus
  3. rRNA and ribosomal proteins are then assembled into the large and small subunits within the nucleolus
  4. The 2 subunits are then transported out of the nucleus into the cytoplasm where they associate to form ribosomes
  5. Ribosomes are the sites of protein synthesis
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7
Q

Structure & role of tRNA

A
  1. tRNA molecule is a single-stranded molecule
  2. A triplet base sequence known as anticodon is present on the tRNA
    -> anticodon forms h bonds with codon on mRNA via cbp
  3. The 3’ end is known as the acceptor stem
    -> site for amino acid attachment
  4. The clover leaf structure of tRNA further folds into a 3-dimensional structure
    -> tRNA is able to fit into the ribosome during translation
  5. tRNA transfers the correct amino acid to the ribosome during translation
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8
Q

Similarities and differences between DNA and RNA

A
  1. Both are polynucleotides made up of nucleotides, which consists of a pentose sugar, nitrogenous base and phosphate group
  2. Polynucleotides have a sugar-phosphate backbone joined by phosphodiester bonds formed by condensation reaction between nucleotides
  3. Both make use of 3 common nitrogenous bases (adenine, guanine & cytosine)
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9
Q

DNA replication

A
  • Unwinding
  • Priming
  • Elongation
  • Termination
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10
Q

DNA replication: unwinding

A
  1. A portion of double helix is unwound and unzipped at the origins of replication by DNA helicase
  2. As helicase moves along the double helix just in front of DNA polymerase, the 2 parental strands separate by breaking h bonds between nitrogenous bases
  3. As the 2 strands separate, a replication bubble is formed with 2 replication forks
  4. DNA replication proceeds in both directions from the origins of replication
  5. For a prokaryotic chromosome, a single origin of replication is present. Eukaryotic chromosomes may have multiple origins of replication
  6. each strand is bound and stabilised by single-stranded binding proteins (ssbp), preventing the strands from rewinding behind the replication fork
  7. DNA topoisomerase introduces a break in a single strand, thus allowing the strand to rotate around the break, and reseals the strand, eliminating the positive supercoil in front of the replication fork
  8. Each parental strand acts as a template for the synthesis of daughter strand
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11
Q

DNA replication: priming

A
  1. Primase binds to the single-stranded DNA template and synthesise RNA primers in the 5’ to 3’ direction
  2. Ribonucleotides are added one at a time via cbp, using the parental DNA strand as a template on both sides of a replication fork
  3. ssbp are displaced where RNA primers are
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12
Q

DNA replication: elongation

A
  1. DNA polymerase adds deoxyribonucleoside triphosphates (dNTP) to the free 3’OH end of the RNA primer
  2. DNA polymerase catalyses the synthesis of a new DNA strand in the 5’ to 3’ direction via cbp, A=T and G_=C with the parental strand
  3. DNA polymerase catalyses the formation of a phosphodiester bond between the 3’OH end of the primer and 5’phophate group of the dNTP added
  4. As the 2 parental strands are antiparallel and polymerisation proceeds in the 5’ to 3’ direction on both sides of the replication fork, one daughter strand is synthesised towards and away from the r fork (leading, lagging strand)
  5. The leading strand is synthesised continuously in the 5’ to 3’ direction towards the replication fork
  6. The lagging strand is synthesised discontinuously, via a series of Okazaki fragments in the 5’ to 3’ direction away from the replication fork
  7. RNA primers are excised and replaced with deoxyribonucleotides by another DNA polymerase
  8. DNA ligase catalyses the formation of phosphodiester bond between 2 Okazaki fragments
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13
Q

DNA replication: termination

A
  1. The product of replication is 2 daughter DNA molecules formed from 1 original parental DNA molecule
  2. Each daughter molecule contains one strand conserved from the parental molecule and one newly synthesised strand
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14
Q

End-replication problem

A
  1. When a eukaryotic linear DNA molecule is replicated, the DNA polymerase is unable to completely replicate to the end of the chromosome
  2. DNA polymerase requires a free 3’OH to add deoxyribonuceotides to
  3. RNA primers at the 5’ end of the newly synthesised strands are excised but cannot be replaced due to the absence of a 3’OH for polymerisation reaction
  4. The newly synthesised strand has a shorter 5’ end due to the removed primers, while the parental strand at the 3’ end is longer

single-stranded 3’ overhang

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15
Q

Transcription: Initiation (prokaryotes)

A
  1. Sigma factor of RNA polymerase recognises and binds to the double-stranded DNA at both the -35 and -10 sequences of promoter
    -> -10 sequence is a consensus sequence of 5’-TATAAT-3’ known as a Pribnow box
  2. Sigma factor is then released from the RNA polymerase
  3. RNA polymerase unwinds and separates the two strands of DNA by breaking h bonds between nitrogenous bases
  4. The template strand is then available for cbp with ribonucleotides
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16
Q

Transcription: Initiation (eukaryotes)

A
  1. TATA binding protein (tbp) recognises and binds to TATA box of promoter, located at 25 base pairs upstream of transcriptional start site
  2. General transcription factors and RNA polymerase are recruited to form the transcription initiation complex
  3. RNA polymerase unwinds and separates the 2 strands of DNA by breaking h bonds between nitrogenous bases
  4. the template strand is them available for cbp with ribonucleotides
17
Q

Transcription: Elongation

A
  1. RNA polymerase adds ribonucleoside triphosphates (NTP) to the free 3’OH end of the growing RNA chain
  2. RNA polymerase catalyses the synthesis of a new RNA strand in the 5’ to 3’ direction via cbp with the template strand
  3. RNA polymerase catalyses the formation of a phosphodiester bond between the 3’OH end of the RNA and 5’phosphate group of the NTP added
18
Q

Transcription: termination (prokaryotes)

A
  1. Termination occurs after a terminator sequence found on the DNA template strand is transcribed
19
Q

Transcription: termination (eukaryotes)

A
  1. Termination occurs after the terminator sequence, polyadenylation signal sequence found on the DNA template strand, is transcribed
  2. It codes for the polyadenylation signal sequence (AAUAAA) in the pre-mRNA
  3. After pre-mRNA is released, post-transcriptional modification occurs before the mature mRNA is transported to the ribosomes in the cytoplasm via the nuclear pore
20
Q

5’ Cap structure

A
  1. Methyl guanosine triphosphate is added to the 1st nucleotide by a 5’–5’ triphosphate linkage. This structure is called a 5’ methylguanosine cap.
  2. Addition of the 5’ methylguanosine cap is catalysed by nuclear enzyme, guanylyl transferase
21
Q

5’ Cap function

A
  1. Facilitate the export of the mature mRNA from the nucleus into the cytoplasm
  2. Protects the mRNA from 5’ exonucleases, hence confers stability to the mRNA
  3. Facilitates the binding of ribosomes to mRNA
22
Q

3’ Poly-A Tail structure

A
  1. About 200 adenine residues are added to the 3’ end of the pre-mRNA called the 3’ poly-A tail
  2. Addition of poly-A tail is catalysed by poly-A polymerase
23
Q

3’ Poly-A Tail function

A
  1. Facilitate the export of the mature mRNA from the nucleus into the cytoplasm.
  2. Slows down degradation by 3’ exonucleases
24
Q

RNA Splicing

A
  1. RNA splicing is the removal of introns and joining of exons together
  2. Introns are removed from the pre-mRNA while exons are ligated together
  3. Small nuclear ribonucleoproteins (snRNPs) are located in the nucleus
  4. Specific snRNPS recognises and binds to the 5’ splice site and 3’ splice site
  5. Additional proteins interact with snRNPs to from spliceosome
  6. Spliceosome excise the introns and join the exons that flanked the introns, releasing introns in a lariat structure
25
Q

Amino Acid Activation

A
  1. Each amino acid is joined to the correct tRNA by a specific enzyme called the aminoacyl-tRNA synthetase
  2. The active site of aminoacyl-tRNA synthetase recognises and binds to a specific pair of amino acid and tRNA
  3. The synthetase catalyses the covalent attachment of an amino acid to the 3’ acceptor stem of the corresponding tRNA
  4. Each tRNA has a specific anticodon which binds to the codon on mRNA by cbp
  5. The resulting aminoacyl tRNA (activated amino acid) is released from the enzyme
26
Q

Translation: Initiation

A
  1. A small ribosomal subunit binds to both 5’ end of mRNA and a specific initiator tRNA, which carries the amino acid methionine, with the aid of initiation factors
  2. The small subunit scans, downstream along the mRNA, until it reaches the start codon AUG. This establishes the start of a reading frame for the mRNA
  3. H bonds are formed between complementary bases of the methionine-tRNA anticodon and start codon on mRNA
  4. Large ribosomal subunits attaches and binds to mRNA, forming the translation initiation complex
  5. At the completion of the initiation complex, initiator tRNA sits in the P (peptidyl) site of the ribosome, and the vacant A (aminoacyl) site is ready for the next aminoacyl tRNA
27
Q

Translation: Elongation

A
  1. The anticodon of an incoming tRNA cbp with the mRNA codon in the A site
  2. Peptidyl transferase catalyses the formation of a peptide bond
  3. Peptide bond is formed between new amino acid at A site and the carboxyl end of the growing polypeptide at the P site
  4. Ribosomes translocate by a codon along the mRNA.
  5. The empty tRNA in the P site is now in the E (exit ) site, where it is released
28
Q

Translation: Termination

A
  1. The base triplets, UAG, UAA, UGA do not code for amino acids but instead signal to stop translation
  2. Release factor binds directly to the stop codon in the A site
  3. Release factor causes addition of a water molecule instead of an amino acid to the polypeptide chain
  4. The reaction hydrolyses the completed polypeptide from the tRNA in the P site

A level bio (11 decks)

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