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Biology > DNA and Protein Synthesis > Flashcards

DNA and Protein Synthesis Flashcards

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AP® Biology flashcards Biology 101 flashcards
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

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

A

Process of producing pre-mRNA using DNA as a template

19
Q

Explain Transcription

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

A

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

21
Q

Reasons for splicing

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

A

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

23
Q

Post-transcriptional modifications

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

A

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

25
Q

Explain Translation

A
  • (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
Q

Difference between genome and proteome

A
  • 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
Q

Describe what is meant by semi conservative DNA replication

A
  • 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
Q

Explain how organic bases help stabilise the structure of DNA

A
  • many hydrogen bonds

- provide strength

29
Q

Suggest role of stop codon

A
  • stop translation

- result in detachment of polypeptide chain from ribsome

30
Q

Suggest why enzyme action is in opposite directions during DNA replication

A
  • 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
Q

Explain how the primary structure of an enzyme determines its three dimensional (tertiary) structure

A
  • sequence of amino acids
  • folded differently
  • bonds form in different places
32
Q

Why does DNA have a stable structure

A
  • 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
Q

Why is it important that bases are joined by weak hydrogen bonds

A
  • 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
Q

Why is base pairing an advantage

A
  • 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
Q

Why is a stable structure for DNA vital

A

DNA must have a stable structure so it can pass from generation to generation without mutation

36
Q

DNA Replication

A
  • 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
Q

State the role of the following enzymes in DNA replication:

  • DNA helicase
  • DNA polymerase
  • DNA ligase
A
  • 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
Q

Purine

A

A and G

Double ring

39
Q

Pyrimidine

A

T and C

Single ring

40
Q

Compare mRNA and tRNA

A
  • 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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