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Biology > DNA and Protein Synthesis > Flashcards

DNA and Protein Synthesis Flashcards

(40 cards)

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1
Q

Gene

A

Section of DNA that codes for a polypeptide chain of functional RNA

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2
Q

Locus

A

Precise location of gene on chromosome

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3
Q

Intron and Exon

A

Intron - non coding

Exon - coding

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4
Q

Unambiguous

A

Each codon only codes for one amino acid

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5
Q

Degenerate

A

Amino acids are coded for by more than one codon

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6
Q

Non overlapping

A

Each base is only part of one codon

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7
Q

Universal

A

Each codon codes for the same amino acid in (almost) all organisms

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8
Q

Start codon

A

Methionine (AUG) is at the start of every polypeptide chain and is removed if not needed

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9
Q

Stop codon

A

Do not code for any amino acid

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10
Q

DNA in Prokaryotic vs Eukaryotic cells

A
  • shorter vs longer
  • circular vs linear
  • not associated with proteins vs histones
  • none vs chromosomes
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11
Q

DNA-histone complex

A

DNA molecule coils around histones which act like spools allows DNA to condense and form nucleosomes

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12
Q

Codon

A

Refers to a sequence of three bases that codes for a single amino acid

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13
Q

Genome

A

Complete set of genes in a cell including those in mitochondria and chloroplasts

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14
Q

Proteome

A

Full range of proteins produced by the genome

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15
Q

rRNA

A

Found in ribosomes

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16
Q

Features of mRNA

A
  • single stranded
  • varies in length depending on gene transcribed
  • template for protein synthesis
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17
Q

tRNA

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A
  • single stranded folded in clover shape
  • small molecule with definite length
  • one end has anticodon to mRNA
  • one end has amino acid attachment site
18
Q

Transcription

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A

Process of producing pre-mRNA using DNA as a template

19
Q

Explain Transcription

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A
  • DNA helicase breaks hydrogen bonds
  • double helix unwinds and separates into two strands
  • RNA polymerase binds to promoter region on template strand (transcribes in 5’ > 3’ direction)
  • complementary base pairing between template strand and RNA nucleotides (U instead of T)
  • terminator sequence stops RNA polymerase
  • mRNA released
20
Q

Splicing

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A

Removal of introns from pre-mRNA using spliceosomes and joining of functional exons

21
Q

Reasons for splicing

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A
  • introns are non coding sections that prevent protein synthesis
  • mRNA would be too large to leave nucleus
  • allows for alternative splicing which means a single genes can code for multiple proteins
22
Q

Suggest why splicing is not necessary in prokaryotic cells

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A

Not necessary since DNA does not contain introns so transcribed directly into mRNA

23
Q

Post-transcriptional modifications

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A

5’ cap and attachment of poly A tail to 3’ end of mRNA to prevent DNA from being degraded by enzymes in nucleus

24
Q

Translation

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A

Process by which a sequence of mRNA forms a polypeptide which a specific amino acid sequence

25
Explain Translation
- (mRNA acts as template) - mRNA binds to ribosome at two binding sites - first tRNA with anticodon binds to start codon carrying with it a specific amino acid - tRNA binds with next codon - enzyme on ribosome (using ATP) catalyses formation of peptide bond between amino acids on first and second tRNA - ribosome moves along mRNA to next codon - first tRNA is released and new tRNA binds - ribosome reaches stop codon and causes release of polypeptide - (many ribosomes bind to the same mRNA)
26
Difference between genome and proteome
- genome is complete set of genetic code where proteome is collection of proteins - genome is relatively unchanging where proteomes are dynamic and change in response to environmental signals - genome is found in nucleus where proteome is found is cytoplasm
27
Describe what is meant by semi conservative DNA replication
- double stranded DNA molecule separates and each strand is used as a template - new DNA is made from one parent strand and one newly synthesised strand
28
Explain how organic bases help stabilise the structure of DNA
- many hydrogen bonds | - provide strength
29
Suggest role of stop codon
- stop translation | - result in detachment of polypeptide chain from ribsome
30
Suggest why enzyme action is in opposite directions during DNA replication
- DNA has antiparallel strands - DNA polymerase has SPECIFIC shape active site - DNA polymerase only complementary to 5' end of template strand - DNA polymerase only complementary to 3' end of developing strand (new strand is synthesised in 5' to 3' direction)
31
Explain how the primary structure of an enzyme determines its three dimensional (tertiary) structure
- sequence of amino acids - folded differently - bonds form in different places
32
Why does DNA have a stable structure
- strong phosphodiester backbone protects bases from corruption by physical or chemical forces - many H bonds between bases (x3 between C-G and x2 between A-T)
33
Why is it important that bases are joined by weak hydrogen bonds
- H bonds can be broken so double-stranded DNA can separate into single strands - to act as templates for DNA replication and transcription (produce mRNA for protein synthesis)
34
Why is base pairing an advantage
- allows DNA to act as template - during semi conservative DNA replication so identical strands can be synthesised - or for transcription to copy gene as mRNA
35
Why is a stable structure for DNA vital
DNA must have a stable structure so it can pass from generation to generation without mutation
36
DNA Replication
- DNA helicase breaks hydrogen bonds - double helix unwinds and separates into two strands - exposed strand acts as template - free complementary nucleotides bind by complementary base pairing - DNA polymerase join nucleotides by phosphodiester bonds to form new complementary strands - semiconservative replication (define) (extra) - DNA polymerase builds leading strand continuously in 5' to 3' direction - lagging strand built discontinuously in Okazaki fragments - DNA lipase joins Okazaki fragments by catalysing formation of phosphodiester bonds
37
State the role of the following enzymes in DNA replication: - DNA helicase - DNA polymerase - DNA ligase
- helicase breaks H bonds to unwind DNA - polymerase adds nucleotides (in 5' to 3' direction) and joins nucleotides by catalysing formation of phosphodiester bond - ligase joins Okazaki fragments by catalysing formation of phosphodiester bond
38
Purine
A and G | Double ring
39
Pyrimidine
T and C | Single ring
40
Compare mRNA and tRNA
- mRNA longer - mRNA straight where tRNA clover shaped - tRNA has H bonding but mRNA does not (do not mention variable length vs definite length)

Biology (14 decks)

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