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Biosciences > Molecules, Cells and Variation - 1.3 + 1.4 > Flashcards

Molecules, Cells and Variation - 1.3 + 1.4 Flashcards

1
Q

Function of sugar-phosphate backbone in DNA?

A

Gives strength

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2
Q

Function of the coiling of DNA?

A

Gives compact shape

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3
Q

Function of the double helix in DNA?

A

Each strand serves as a template in replication.
Protects genetic code (bases)
Makes molecule more stable

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4
Q

How does DNA being large support its function?

A

Allows large amount of information to be stored.

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5
Q

How does the high number of hydrogen bonds in DNA benefit its function?

A

Gives stability
Prevents code being disrupted
Allows chain to unzip easily for replication + transcription

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6
Q

How does the sequence of bases in DNA benefit the function?

A

Provides genetic code for protein synthesis

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7
Q

How does complementary base pairing benefit the function of DNA?

A

Enables information to be replicated accurately

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8
Q

DNA chromosome structure in eukaryotes

A

DNA exists in the nucleus, surrounded by membrane.
Histones associate with a region of DNA to form a nucleosome. DNA is further coiled making chromatin.
DNA is present as indistinct chromatin in the nucleus for most of the cell cycle. During mitosis, DNA condenses and coils further, becoming chromosomes.

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9
Q

Chromatin

A

Structure formed when DNA is packaged around histones and super coiled. Provides a compact store of genetic information.

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10
Q

Nucleosome

A

Subunit of chromatin, consisting of sections of DNA wrapped around histones. Provides structural support.

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11
Q

DNA structure in prokaryotes

A

DNA is smaller, circular and unassociated with proteins. No defined nucleus. During replication, DNA attaches to mesosome.

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12
Q

Semi-Conservative method of DNA replication

A

1) Strands separate as hydrogen bonds break. Promoted by helicase.
2) Strand act as templates for formation of new complementary strands.
3) DNA nucleotides align next to template strands according to their specific base pairing.
4) Nucleotides join forming a polynucleotide strand using DNA polymerase. Hydrogen bonds reform.
5) Two DNA molecules are identical to each other and to the original DNA.

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13
Q

Why is DNA replication semi-conservative?

A

Each newly formed DNA molecule contains one of the original polynucleotide strands and one newly synthesised from new individual nucleotides.

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14
Q

Why does the 5’, 3’ direction of synthesis in DNA make replication more complicated?

A

DNA polymerase can only work in the 3’, 5’ direction. The 3’, 5’ strand is replicated in it’s entirety and is the leading strand, allowing continuous replication.
The 5’, 3’ strand is formed in smaller fragments, making the 3’ available. DNA ligase has to join the fragments together, causing discontinuous replication.

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15
Q

How does RNA differ in structure to DNA?

A
  • Pentose sugar is Ribose.
  • Uracil replaces Thymine
  • Are single stranded
  • Shorter in length, with a lower molecular weight
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16
Q

Ribosomal RNA

A
  • Makes up 80% of the total RNA of the cell. Is synthesised on genes present on DNA.
  • rRNA synthesis occurs in the nucleolus.
  • It moves to cytoplasm via nuclear pores, it then associates w/ protein molecules, forming ribosomes.
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17
Q

Messenger RNA

A
  • Formed from a single DNA strand during transcription.
  • Has a variable length depending on gene.
  • Mostly exists for short time. Degrades after protein synthesis.
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18
Q

Transfer RNA

A
  • Acts as intermediate molecule between triplet code of mRNA and amino acid sequence present in polypeptide.
  • It transfers amino acids in cytoplasm to ribosome.
  • All have same basic structure. Clover leaf shape.
  • 5’-end of tRNA always ends in guanine, 3’-end always ends in CCA.
  • Anticodon is situated on bottom loop. Directly related to amino acid carried by tRNA.
  • Rest of molecule has variable base sequence.
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19
Q

Gene

A

Region of DNA, whose nucleotide base sequence codes for the production of a specific polypeptide/protein. Chromosome may contain many hundreds of genes.

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20
Q

Locus

A

Position of the gene on the DNA

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21
Q

Allele

A

Different forms of a gene, which code for similar polypeptides, and are located on the a similar locus on homologous chromosomes. They carry genes controlling the same characteristics but not necessarily the same alleles.

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22
Q

Transcription

A

1) DNA uncoils and strands separate.
2) One DNA strand acts as a template strand (non-coding sequence).
3) Single RNA nucleotides line up alongside the template strand (specific complementary base pairing).
4) RNA polymerase joins nucleotides together forming backbone of pre-mRNA.
5) DNA strands in nucleus recoil when mRNA has been produced.

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23
Q

Splicing

A

Spliceosome removes introns and joins exons together to form mRNA. This leaves the nucleus and attaches to a ribosome allowing translation.
Prokaryotes don’t have introns.

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24
Q

Initiation of translation

A

1) Ribosome moves along mRNA and finds start codon. 1st tRNA binds at P site on ribosome.
2) tRNA’s anticodon pairs w/ specifc codon on mRNA .
3) 2nd tRNA binds at A site.
4) Bond between 1st amino acids and tRNA is broken. Released energy forms peptide bond between 1st + 2nd amino acid.
5) Dipeptide attached to second tRNA has formed.
6) Disassociated tRNA moves away to gain new amino acid.
7) Dipeptide attached to 2nd tRNA moves to P site.

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25
Q

Elongation of translation

A

1) Peptide chain continues to grow and ribosome moves along to next new codon.
2) New codon is aligned with A site, allowing the next tRNA to interact with the ribosome.
3) Dipeptide is linked to 3rd amino acid, 2nd tRNA moves away to amino acid pool in cytoplasm.
4) Process continues along mRNA strand until all codons have been ‘read’ and polypeptide has been produced.

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26
Q

Termination of translation

A

1) When a stop codon is reached, bond between polypeptide and last tRNA is hydrolysed. Ribosome leaves mRNA and can reattach at the start again.
2) As the protein is synthesised it can develop different structures. May undergo further modification in the ER if exported out the cell.

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27
Q

Regulation of transcription

A
  • Gene must be stimulated by transcriptional factors, -
    These bind to DNA upstream from specific gene, and stimulate RNA polymerase to begin transcription, allowing protein synthesis.
  • An inhibitor prevents transcription factors from binding when a gene is not expressed.
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28
Q

Regulation of translation

A
  • Ccan be prevented by breaking down mRNA before it is translated at the ribosome.
  • Small interfering RNA (siRNA) carries this out.
  • RNA is cut into smaller pieces forming siRNA.
  • One strand combines with an endonuclease enzyme.
  • siRNA binds via complementary base pairing.
  • Signalling endonuclease to cut mRNA where siRNA binds.
  • mRNA is now unable to be translated.
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29
Q

Epigenetics

A

Refers to a change in phenotype without a change in genotype.

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30
Q

Example of changes to gene expression

A
  • Methylation of DNA
  • Acetylation of histone tails
  • Phosphorylation of amino acids
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31
Q

Gene mutation

A
  • Changes in the nucleotide base sequence of DNA.
  • Results in formation of a different polypeptide.
  • Altered base sequence codes for different amino acid sequence. Produces new alleles.
  • Occur naturally at random, rate varies.
  • Can arise due to incorrect pairing in DNA replication.
  • Most are harmful and recessive.
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32
Q

Mutagenic agent examples

A
  • High energy radiation e.g. X-rays, gamma rays, U.V. light.
  • High energy particles e.g. alpha and beta particles.
  • Chemicals such as nitrous oxide or benzene
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33
Q

Substitution as a gene mutation

A
  • Replacement of one or more bases by another.
  • Same amino acid may be coded for, (degenerate), so polypeptide remains unchanged.
  • Could be one different amino acid, but a functional protein is still produced.
  • Different amino acid is coded for, creating a non-functional protein.
  • Could form a stop codon, terminating polypeptide, creating a non-functional protein.
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34
Q

Deletion as a gene mutation

A
  • Removal of one or more bases.
  • May result in a frame shift. All future codons change.
  • Amino acid sequence is altered, non-functional protein formed.
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35
Q

Addition as a gene mutation

A
  • Addition of one or more bases.
  • Leads to a frame shift.
  • Sequence of amino acids is altered, non-functional protein formed.
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36
Q

How can a gene mutation affect the activity of an enzyme?

A
  • Active site may change, so substrate can’t attach making the enzyme non-functional.
  • Slow down enzyme activity.
  • Affects tertiary structure and shape of active site.
  • Can block a metabolic pathway.
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37
Q

Genetic Engineering

A
  • Genes taken from an organism are inserted into another, altering the genetic make-up. Making a transgenic organism.
  • Microorganisms are often used as recipient cells during gene transfer. Phages are also used.
  • Rapid reproduction of microorganisms allows fast production of gene product.
38
Q

In vivo cloning

A

Genetic copies are made inside a living organism

39
Q

In vitro cloning

A

Genetic copies are made outside a living organism

40
Q

Reverse transcriptase

A

Enzyme used to generate complementary DNA from RNA, typically found in retroviruses.

41
Q

Using reverse transcriptase in gene cloning

A
  • mRNA is used as a template for required protein/gene
  • mRNA is mixed w/ free DNA nucleotides and enzyme
  • Free nucleotides align next to their complementary bases on the mRNA template
  • Enzyme then joins nucleotides together
  • cDNA is produced which can then form DNA using DNA polymerase
42
Q

Restriction endonuclease

A

Enzyme used to cleave DNA into fragments at specific recognition sites on the molecule. Found in bacteria. Cleaves sugar-phosphate backbone resulting in sticky ends or blunt ends to DNA.

43
Q

Palindromic in terms of recognition sites

A

Read the same either way on each polynucleotide strand

44
Q

Why are ‘sticky ends’ beneficial in gene cloning?

A

Enable DNA to be joined or spliced onto a different piece of DNA more effectively due to the staggered cut. Preferred.

45
Q

Why are ‘blunt ends’ less favourable in gene cloning?

A

More chance of the ends joining non-specifically

46
Q

DNA ligase in relation to gene cloning

A
  • Specific DNA can be inserted into plasmid DNA and then permanently joined together using DNA ligase. This reforms the sugar-phosphate backbone by reforming phosphodiester bonds between nucleotides.
  • New recombinant DNA can now be replicated by host organism, or copied via an artificial process like PCR .
47
Q

Competent in relation to gene cloning

A

Primed to take up DNA more readily

48
Q

What type of change in culture conditions for prokaryotes would mean they more readily took up a DNA fragment>

A

A high calcium ion concentration alters membrane properties, making it more permeable.

49
Q

Transfer of genes

A
  • Plasmids/vectors containing the recombinant DNA are added to competent bacteria, endocytosis can take up plasmid
  • If a bacterium takes up the plasmid it is said to be transformed and is now a transgenic organism.
  • Viruses, liposomes and ballistic projectiles can be used as vectors to allow plasmids to be taken up by cells and tissue
50
Q

Electroporation

A
  • More efficient method of gene transfer.
  • Bacteria + plasmids are mixed together; high voltage is applied for short period
  • disruption of cell membrane allows DNA to pass into the cell
51
Q

Why are gene markers necessary in gene cloning?

A
  • Transformation of bacteria is not very efficient. Many bacteria either won’t pick up a plasmid or the plasmid won’t incorporate the foreign gene.
  • Marker genes enable genetically engineered bacteria w/ a recombiant plasmid to be detected and isolated for subsequent culturing
52
Q

Antibiotic resistance marker genes

A
  • Genes present in some plasmids can be used as markers. E.g. genes for ampicillin and tetracycline.
  • Inserted gene will disrupt resistance gene already present, so resistance is lost.
  • Transformed bacteria can be identified as they will be resistant to antibiotics, e.g. ampicillin, not tetracycline
53
Q

Fluorescent marker genes

A
  • Used in place of antibiotic resistance genes due to risk of spread of antibiotic resistance by horizontal gene transmission from one species to another.
  • GFP gene codes for production of a green fluorescent protein (GFP). Can be easily identified w/ UV light.
54
Q

Obtaining gene products

A
  • Bacteria which take up recombinant plasmids, replicate w/ it, and produce a colony of clones.
  • This can be harvested and stored.
  • Plasmid can be purified, digested with restriction enzymes and analysed using gel electrophoresis to confirm the presence and size of foreign DNA.
  • Recombiant bacteria are cultured and produce gene product in large amounts
55
Q

Large scale culturing

A
  • Industrial fermenters can culture bacteria large scale.
  • Chemicals made: enzymes, antibiotics + hormones.
  • Fermenters are sterilised before adding liquid medium and GE bacteria.
  • Optimum growth conditions are ensured.
  • Colony grows + produce gene product which can be extracted and purified
56
Q

Electrophoresis

A

Technique used to separate macromolecules, by a gel matrix of agarose/polyacrylamide. Gel acts as a sieve, w/ larger molecules move slower and smaller ones move faster.

57
Q

Agarose

A

Polysaccharide from seaweed

58
Q

Process of electrophoresis

A
  • DNA samples are mixed with a loading buffer and placed into wells.
  • Current is applied
  • As DNA moves through gel, it separates according to size.
  • Staining w/ a fluorescent dye and using UV light allows visualization of fragments.
  • Gel can be photographed to record position + fragments of DNA can be cut out + extracted from gel.
  • Size is found by comparison w/ a DNA ladder
59
Q

Loading buffer

A

Sucrose and blue dye, shows progression of separation.

60
Q

Why is a current applied in electrophoresis?

A

Negative current applied which repels nucleic acids due to their similarly negative charge. They migrate through the gel towards the positive anode.

61
Q

DNA ladder

A

Scale showing DNA of known size which can be compared to a gel electrophoresis, to measure length of unknown DNA.

62
Q

Gene probes

A
  • Short single stranded lengths of DNA.
  • Can be used in various ways to detect particular DNA sequences.
  • Can be used to identify genes w/ a target sequence. -
  • Can identify mutations.
  • Used in genetic fingerprinting
63
Q

Disadvantages of genetic engineering

A
  • Transfer of genes to non-target organisms, resulting in disruption of normal functions.
  • Irreversible process, w/ no definite economic benefits.
  • Ethics regarding altering genetic make up of animals and destruction of embryos.
  • Unknown ecological + evolutionary consequences
  • Development of new resistant species
  • Consumption of GE food w/ foreign proteins may produce an immune response.
  • Accidental transfer of unwanted genes by vector
64
Q

Separation and identification in DNA sequencing

A
  • Products from reaction tubes are placed in wells.
  • Fragments from reaction tubes are separated by acrylamide gel electrophoresis
  • DNA fragments are separated according to size
  • Gel containing separated DNA fragments are transferred to a sheet of filter paper, and analysed by autoradiography
  • Gel is read bottom up, to determine DNA sequence
  • Sequence is complementary to copied strand.
65
Q

Benefit of acrylamide gel electrophoresis

A

Can separate DNA due to a single base pair difference in length

66
Q

How are gene probes used?

A
  • 15-20 bases long, typically radioactive due to P-32 being present in sugar-phosphate backbone.
  • Relies on target sequence being known, allowing complementary base pairing.
  • Detection of pairing occurs either by blotting technique or DNA is labelled instead (e.g. UV)
67
Q

Genetic fingerprinting

A

Analysis of introns in DNA, where mutations often occur throughout and can repeat. Called VNTRs.

68
Q

VNTRs

A

Variable number tandem repeats. Identifies variation in population by number of repeats in a DNA sequence. Every individual has 2 alleles of VNTRs of the same locus, from each parent.

69
Q

RFLP

A
  • RF = Restriction Fragments. DNA fragments cut by restriction enzymes.
  • L = Length, refers to length of restriction fragments.
  • P = Polymorphism, means many shapes. Lengths of restriction fragments differ greatly, due to VNTR’s, so DNA length differs throughout population.
70
Q

Procedure of genetic fingerprinting

A
  • DNA is purified from an individual.
  • Restriction enzymes cut at sites flanking the VNTRs
  • Southern blotting method is carried out
  • If a single probe is used a maximum of two bands may be detected (2 alleles). 2 bands mean Aa, 1 = AA
  • If multiple probes are used for different sequences, multiple bands are seen. Increasing reliability of data as it allows easier distinguishing between 2 individuals.
  • Autoradiography used to detect probe position on blot. Fingerprint is produced.
71
Q

Fingerprint in genetic fingerprinting

A

Unique genetic profile

72
Q

PCR

A

Technique used to amplify a specific nucleic acid sequence in vitro in order for other methods to be used to determine its size, fingerprint, or nucleotide sequence etc

73
Q

How many PCR be used in forensic science?

A
  • Biological material containing DNA (e.g. blood, semen, follicle, cheek cells, bone) can be used to characterise the victim/criminal using very little DNA. Recent.
  • Police can now solve old cases using modern methods w/ stored material
74
Q

How is PCR used in phylogenetic studies?

A

DNA sequence information may be obtained from old/extinct biological specimens allowing evolutionary link to be investigated

75
Q

How is PCR used in blood screening and blood product screening?

A
  • Blood and products can be screened to detect contamination by viruses such as HIV, Hepatitis, CMV.
  • This prevents transfusions being contaminated and the recipient becoming infected
76
Q

How can PCR be used in patient monitoring?

A
  • Success of antiviral treatments in immuno-compromised individuals (chemotherapy patients’ AIDS patients) can be followed.
  • DNA produced can be used in gene therapy
77
Q

How is PCR used in genetic screening?

A
  • Info about diseases is being collected. Mainly inheritable ones. Allows those at risk to plan their life.
  • W/ results of human genome project, info about a genetic component marking an individual as high or low risk could be determined.
  • Allows amplification of an isolated gene.
78
Q

Stage 1 of PCR

A

Separation of strands

  • Reaction mix has primers, template DNA, nucleotides and DNA polymerase
  • Mixture is heated to 95°C for 2-5mins. Breaks hydrogen bonds and separates DNA.
79
Q

Stage 2 of PCR

A

Binding of primers

  • Mixture is cooled to 45-65°C (T depends on length and sequence of primers)
  • Primers bind to target sites on DNA, indicating sections to be copied.
  • DNA polymerase binds to primer/DNA.
80
Q

Stage 3 of PCR

A

Copying of DNA

  • Temperature raised to 72°C (enzyme optimum)
  • Complementary nucleotides base pair w/ template strand
  • DNA polymerase moves along strand and joins them together, forming new sugar-phosphate backbone
  • 2 new double-strands are produced
81
Q

Stage 4 of PCR

A

Repeat

  • Cycle is repeated 20-40 times causing exponential increase.
  • Mixture analysed by gel electrophoresis
82
Q

DNA sequencing

A
  • Technique is used to determine sequence of nucleotides in DNA sample. May have been identified + obtained from a Southern blot + inserted into a plasmid, then inserted into bacteria for replication. The plasmid then being purified and used as a source of DNA to be sequenced.
  • Most common technique is the Sanger procedure, which requires four sequencing reactions to be carried out at the same time. Needs a large amount of DNA.
83
Q

What does a DNA sequencing reaction tube contain?

A
  • Large DNA quantities (templates)
  • Radioactively labeled DNA primers
  • Nucleotides (complementary base pairing)
  • DNA polymerase (makes new strand)
  • Terminator nucleotides
84
Q

Why are radioactively labelled DNA primers present in DNA sequencing reaction tube?

A
  • Primer molecules are used to start sequencing

- Radioactivity allows presence of primer to be detected. Alternatively, DNA can be labelled.

85
Q

Why are terminatory nucleotides present in DNA sequencing reaction tube?

A
  • Small percentage of each one of the four DNA nucleotides is modified. Acts as a terminator nucleotide in the extending DNA strand.
  • Nucleotide is a dideoxy nucleotide. Terminating synthesis.
86
Q

How does a dideoxy nucleotide differ to a deoxyribose nucleotide and how is it relevant w/ DNA replication?

A

DNA polymerase can only synthesise DNA in the 5’ to 3’ direction. Dideoxy nucleotide has a 5’ OH that can attach to the growing nucleotide strand, but it does not have an OH group attached to the carbon 3’ as a result it cannot be extended further.

87
Q

DNA sequencing reaction procedure

A
  • DNA polymerase uses the radioactive primer to begin the formation of a complementary DNA strand.
  • DNA nucleotides are joined together using DNA sample as template.
  • Random addition of a dideoxy nucleotide stops synthesis of DNA strand; majority of reactions continue, and may be terminated later.
  • Due to random termination, one tube will contain different sized DNA fragments (strands) as the DNA molecules are terminated by a specific nucleotide at different positions. All fragments are radioactive due to primer.
88
Q

Molecule involved in regulation of transcription

A

Transcription factors

89
Q

Molecule involved in regulation of translation

A

SiRNA

90
Q

Hierarchy

A

Groups within groups with no overlapping

91
Q

Phylogenic group

A

Grouped according to evolutionary links and relationships

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