Necleotides And Nucleic Acids Flashcards

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1
Q

purines

A

Two carbon ringed bases

Adenine

Guanine

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2
Q

Pyrimidines

A

Single carbon ring bases

Thymine

Cytosine

Uracil (in RNA)

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3
Q

Describe formation of DNA (6)

A

Nucleotides linked together by condensation

Double helix; antiparallel strands

Complementary base Pairs held by weak hydrogen bonds

Phosphodiester bonds between phosphate and pentose

5’ at one end, 3’ at the other

5’ to 3’ leading
3’ to 5 lagging

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4
Q

How is DNA structure suites for its role (3)

A

Polymer; a lot of information stored

Bass sequence; allows coding for information

Double stranded; accurate replication and makes the molecule stable

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5
Q

State the three theorised methods of DNA replication

A

Conservative

Semi-conservative

Dispersive

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6
Q

Semi-conservative replication, explain why it is the most efficient form of replication?

A

Original stand serves as template for a new complementary strand

50% old strand, 50% new

  • Allows few errors to be made as original DNA acts as a template.
  • Errors can be more easily corrected as original strand is still present
  • New strand would see errors more likely to be ignored.
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7
Q

Name the enzymes involved in replication

A

DNA Helicase

DNA polymerase

DNA ligase

DNA Primase

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8
Q

DNA Helicase

A

Separates DNA strands by cutting through HBonda

Uses energy from ATP hydrolysis

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9
Q

DNA Polymerase

A

Catalyses the formation of phosphodiester bonds between nucleotides

Therefore causing deoxyribonucleotides to form DNA strands

Can only move from 5’ to 3’

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10
Q

DNA ligase

A

Links two DNA strands with double strand break

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11
Q

Structure of nucleotide

A

Phosphate links at 5th C on sugar forms- also links with OH group at 3rd C

Pentose sugar( ribose or deoxyribose )

Base; purine or pyrimidines

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12
Q

Okazaki fragments

A

Short piece of DNA created on lagging strand

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13
Q

Difference between continuous and discontinuous replication

A

Continuous;
Moves from 5’ to 3’
DNA polymerase binds to the end
Free nucleotides are added with no breaks

Discontinuous;
3’ to 5’ so DNA cannot bind to end
Free DNA added in sections
Forms Okazaki fragments which are sealed by DNA ligase

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14
Q

Why does DNA not catalyse the joining of Okazaki fragments

A

Enzymes are substrate specific

Nucleotides have a different shape on Okazako fragments compared to normal nucleotides

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15
Q

How many different types of codons are possible

A

64

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16
Q

What is the start codon

A

AUG

17
Q

What are the 3 termination codons

A

UAA

UAG

UGA

18
Q

Triplet(codon)

A

Three consecutive nucleotides

19
Q

Gene

A

Chain of codons with a start and stop.

20
Q

Transcription unit

A

Genes collectively coding for an end product

21
Q

Briefly describe DNA replication

A
  • DNA unzips via DNA Helicase

New DNA strands formed by free nucleotides forming complementary bonds

DNA polymerase adds nucleotides from 5’ to 3’ on leading strand

3’ to 5’ on lagging strand via Okazaki fragments

Stands rewinds together

22
Q

Antisense

A

3’ to 5’ strand during transcription

Codes for mRNA therefore complementary to
mRNA and Sense strand

Contains anti-codons which code for the same as antisense

23
Q

Sense strand

A

5’ to 3’ strand during transcription

Codons code for same amino acids as mRNA

Complementary to antisense

Codes for protein but not used in proteins synthesis

24
Q

Anticodon

A

Attaches to tRNA

Complementary to mRNA codons

25
Q

Ribosome in Translation

A

Large and small subunits

Almost same amount of protein

Forms rRNA

Small subunits bind to mRNA during translation

26
Q

rRNA

A

Maintains structural stability of protein synthesis sequence

27
Q

tRNA

A

Folded to allow attachment of anti codon

Carries amino acid according to anticodon

28
Q

What structure does the sequence of amino acids form

A

Primary

29
Q

Describe transcription (6)

A
  1. ) DNA unwinds and unzips using DNA helicase to break hydrogen bonds between base pairs.
  2. ) Strands separate into sense (5’ to 3’) and antisense (3’ to 5’) strands.
  3. ) Antisense strand acts as a template for the free complementary RNA nucleotides to bind to.
  4. ) RNA Polymerase form phosphodiester bonds between RNA nucleotides until the gene ends to form mRNA
  5. ) mRNA detaches from DNA. DNA rewinds into helical structure.
  6. ) mRNA leaves the nucleus via nuclear pore, into the cytoplasm.
30
Q

Describe translation

A
  1. ) mRNA binds to the smaller subunit in the ribosome at start codon (AUG)
  2. ) Complementary tRNAs bring amino acids according to the anticodons
  3. ) Maximum of 2 tRNAs can be bound at the same time
  4. ) Chain of amino acids form a polypeptide which is primary structure of protein
  5. ) Polypeptide chain stops at ‘STOP’ codons; UGA, UAA, UAG.
  6. ) Polypeptide chain folds to form secondary and tertiary structures.
31
Q

How is the structure of DNA suited for replication (5) and

A
  • Double stranded with each strand acting as a template
  • Complementary base pairs
  • H bonds between base pairs can break easily
  • purine only bond to pyrimidines
  • different hydrogen bonds between AT and GC