Cellular Control Flashcards

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1
Q

Define the term “mutation” (umbrella term).

A

A change in genetic material which may effect the phenotype of the organism.

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2
Q

Define the term “point mutation”.

A

Changes one base on the triplet codon. May or may not have damaging effects.

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3
Q

Define the term “substitution”.

A

A mutation where one or more nucleotides are substituted for another in the DNA strands.

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4
Q

Define the term “deletion”.

A

Mutations where one or more nucleotides are deleted and lost from the DNA strands.

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5
Q

Define the term “frame shift”.

A

The addition or deletion of a base in a sequence. Causes the whole sequence to shift out of place from the point of addition or deletion, so it is read out of sync. Most damaging.

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6
Q

Define the term “silent mutation”.

A

A mutation that occurs in the part of the gene that does not code for a protein (the introns). Therefore, it will not effect the phenotype.

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7
Q

Define the term “missense mutation”.

A

Missense mutatuions result in the incorporation of an incorrect amino acid into the primary structure when the protein is synthesised. Mutation could be silent, beneficial or harmful.

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8
Q

Define the term “nonsense mutation”.

A

Can result in codon becoming a stop codon instead of coding for an amino acid. The result is a shortened protein being synthesised which is normally non-functional.
These mutations tend to have negative/harmful effects on the phenotype.

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9
Q

Define the term “gene mutation”.

A

Change in the base sequence which codes for a polypeptide chain/ protein.

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10
Q

Define the term “chromosome mutation”.

A

Affect the whole chromosome or a number of chromosomes within a cell.

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11
Q

Explain why a change in the sequence of nucleotides of a gene can affect the function of the protein produced from that gene.

A

A change in the base sequence might change the amino acid that a codon codes for. This would disrupt protein synthesis and therefore the proteins expressed in the phenotype. Potentially very harmful effect i.e. if the active site of an enzyme was not synthesised properly.

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12
Q

Describe how a mutation can have a neutral effect, a harmful effect or a beneficial effect, and give an example of each.

A

1) Neutral - Normally function proteins are still synthesised. The mutation occurred in the a part of the gene that does not code for a protein.
2) Damaging - the phenotype of an organism is affected in a negative way because proteins are no longer synthesised or the proteins synthesised are non-functional. Can interfere with essential processes.
3) Beneficial - Very rare. A protein is synthesised that results in a new and useful characteristic in the phenotype. i.e. immunity to HIV.

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13
Q

State the 3 types of mutagen and give an example of each.

A

1) Chemical - deaminating agents - chemically alter bases in DNA.
2) Physical - ionizing radiations (x-ray) - can break one or both of the DNA strands.
3) Biological agents - alkylating agents - methyl or ethyl groups are attached to bases resulting in incorrect pairing of bases during replication.
- base analogs - incorporated into DNA in place of the usual base during replication, changing the base sequence.
- viruses - viral DNA may insert itself into a genome, changing the base sequence.

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14
Q

Name and describe the 4 types of chromosome mutation.

A

1) Deletion - a section of chromosome breaks off and is lost within the cell.
2) Duplication - sections get duplicated on a chromosome.
3) Translocation - a section of one chromosome breaks off and joins another non-homologous chromosome.
4) Inversion - a section of chromosome breaks off, is reversed and then joins back onto the chromosome.

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15
Q

Describe and explain the possible effects of a substitution mutation.

A

Would effect the amino acid that a codon codes for as a base in the triplet codon has been changed. This will effect the protein that is synthesised and possibly its functionality. Does not cause a frameshift.

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16
Q

Describe and explain the possible effects of insertion or deletion mutations.

A

Leads to a frameshift mutation. Every successive codon will be changed after the point of deletion/ insertion. The sequence is therefore read out of sync, the amino acid that each codon codes for being affected. Protein will be synthesised incorrectly.

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17
Q

Define the term “gene expression”.

A

Information from a gene is used in the synthesis of a functional gene product. Often proteins.

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18
Q

Define the term “epigenetics”.

A

The control of gene expression by modification of the DNA. – External control of gene expression.

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19
Q

Name and describe the four levels at which genes (or proteins) are regulated.

A

1) Transcriptional - genes can be turned off and on.
2) Post-transcriptional - mRNA can be modified which regulates translation and the types of proteins produced.
3) Translational - translation can be stopped or started.
4) Post-translational - proteins can be modified after which changes their functions.

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20
Q

Define the term “chromatin”.

A

Un-condensed DNA in a complex with histones. Exists like this in nucleus before cell division.

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21
Q

Define the term “heterochromatin”.

A

Tightly packed DNA during cell division.

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22
Q

Define the term “euchromatin”.

A

Loosely packed DNA during interphase.

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23
Q

Describe how chromatin remodelling allows the expression of some genes but not others.

A
  • The transcription of genes is not possible when DNA is tightly wound (heterochromatin) because RNA polymerase cannot access the genes. However, genes in loosely packed DNA (eurochromatin) can be freely transcribed.
  • You don’t want protein sysnthesis to take place during cell division, and it can’t because heterochromatin is tightly wound. However, protein synthesis can take place at interphase because eurochromatin is loosely wound.
    Simple form of regulation.
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24
Q

Describe how histone modification can affect gene expression.

A
  • Histones can be modified to increase or decrease the degree of packing.
  • The addition of acetyl groups (acetylation) or phosphate groups (phosphorylation) reduces the positive charge on histones (and because DNA is negatively charged) it makes DNA to coil less tightly, allowing certain genes to be transcribed.
  • The addition of methyl groups (methylation) makes the histones more hydrophobic so they bind more tightly to each other. DNA coils more tightly and prevents transcription of genes.
25
Q

Define the term “operon”.

A

A group of genes that are under the control of the same regulatory mechanism and are expressed at the time.

26
Q

Draw and label a diagram to show the lac operon and its associated regulatory gene.

A

1) Regulatory gene - a gene that codes for repressor proteins.
2) CAP binding site -
3) Promoter - site of RNA polymerase binding. RNA polymerase will catalyse the formation of phosphodiester bonds between nucleotides.
4) Operator - repressor protein binding site. RNA polymerase will be blocked if a repressor protein binds here.
5) Structural genes - These are transcribed into mRNA and then translated into proteins. They code for the structural genes needed to metabolise lactose.

27
Q

Define the term “structural gene”.

A
  • Genes that code for structural proteins or enzymes not involved in DNA regulation.
28
Q

Name the proteins produced from the structural genes in the lac operon.

A
  • Lac Z, beta galactosidase, breaks down glucose.
  • Lac Y, lactose permease, brings lactose into the cell.
  • Lac A, transacetylase, unknown function.
29
Q

Describe the roles of the regulatory gene, the structural genes, the operator region and the promotor region of the lac operon for the metabolism of lactose.

A
  • The regulatory gene (lac I) codes for a repressor protein that prevents the transcription of structural genes.
  • The structural genes code for three enzymes that involved in the metabolism of glucose. They are transcribed onto a single long molecule of mRNA but are translated into separate proteins.
  • RNA polymerase binds the promoter, repressor proteins bind to operator.
30
Q

When is lactose used as a respiratory substrate?

A
  • When glucose is in short supply (glucose is more easily metabolised than lactose).
  • Different enzymes are required to metabolise lactose.
31
Q

Explain how the lac operon works when lactose is present in the growth medium.

A
  • The repressor protein is constantly being produced and binds to the operator.
  • Lactose acts as the inducer. When it is present it binds to the repressor protein and changes its shape, this means that it can no longer bind to the operator.
  • This means RNA polymerase is not blocked from binding to the promoter, so the three structural genes are transcribed and the necessary enzymes are synthesised.
32
Q

Explain how the lac operon works when lactose is absent from the growth medium.

A
  • Regulatory genes codes for transcription of repressor proteins.
  • In the absence of lactose, the repressor protein is able to bind to the operator and prevent RNA polymerase from binding to the promoter.
  • This means the structural genes that code for the enzymes that metabolise lactose are not synthesised in the absence of lactose.
  • Saves energy.
33
Q

Describe the role of cAMP in control of the lac operon.

A
  • Binding of RNA polymerase results in a relatively low rate of transcription that needs to be increased in order to synthesise the quantity of enzymes necessary to metabolise lactose.
  • The binding of a cAMP receptor protein (CRP) achieves this increase. However, CRP can only bind to lac operon when it is bound to cAMP (a second messenger molecule).
  • The transport of glucose into the E.Coli cell decreases the levels of cAMP and therefore the transcription of the structural genes that metabolise lactose is decreased.
  • If both glucose and lactose are present, it will still be glucose, the preferred respiratory substrate, that is metabolised.
34
Q

Define the term “intron”.

A

Regions of non-coding DNA or RNA.

35
Q

Define the term “exon”.

A

Regions of coding DNA or RNA.

36
Q

Define the term “pre-mRNA”.

A

The mRNA transcribed from the DNA before any post-transcriptional regulation to remove introns.

37
Q

Define the term “mature mRNA”.

A

mRNA after the removal of introns and any other post-transcriptional changes.

38
Q

Define the term “RNA processing”.

A

Any modification made to RNA between its transcription and its final function in the cell.

39
Q

Define the term “RNA editing”.

A
  • Base sequenced changed through the addition deletion or substitution of a base.
  • Results in the synthesis of different proteins and increases the range of proteins that can be produce from a single RNA molecule or gene.
  • A type of RNA processing.
40
Q

Describe how pre-mRNA is modified to produce mature mRNA, where this occurs and the benefit of this modification process

A
  • Pre-mRNA must be modified to mature mRNA before it can bind to the ribosome for protein synthesis.
  • A cap (modified nucleotide) is added to the 5’ end and a tail (a long chain of adenine nucleotides) is added to the 3’ end.
  • These both help to stabilise mRNA and delay degradation in the cytoplasm. Cap also aids binding mRNA to ribosomes.
  • Splicing occurs where the RNA is cut at specific points and the introns are removed and the exons are joined together.
  • Both processes occur within the nucleus.
41
Q

Describe 3 mechanisms that can regulate protein synthesis at the translational level.

A

1) Degradation of mRNA - the more resistant the molecule the longer it will last in the cytoplasm, that is, a greater quantity of protein synthesised.
2) Binding of inhibitory proteins to mRNA prevents it binding to ribosomes and therefore the synthesis of proteins.
3) Activation of initiation factors which aid the binding of mRNA to ribosomes.

42
Q

Describe the role of protein kinases in regulation of gene expression or protein activity.

A
  • Protein kinases are enzymes that catalyse the addition of phosphate groups to proteins.
  • The addition of a phosphate group changes the tertiary structure and so the function of a protein.
  • Many enzymes are activated by phosphorylation so protein kinases are therefore important regulators of cell activity.
43
Q

Describe 4 ways in which proteins can be modified to provide post-translational control.

A

Modifications to proteins that have been synthesised. This includes:

  • addition of non-protein groups such as carbohydrate chains, lipids or phosphates.
  • modifying amino acids and the formation of bonds such as disulphide bridges.
  • folding or shortening of proteins.
  • modification by cAMP e.g. in the lac operon when cAMP binds the CRP to increase rate of transcription.
44
Q

Define the term “body plan”.

A

The general structure each individual organism assumes as it develops.

45
Q

Define the term “morphogenesis”.

A

The regulation of the pattern of anatomical development.

46
Q

Define the term “homeobox gene”.

A

Homeobox genes are 180 base pairs long and are responsible for the development of body plans. They code for the homeodomain of a protein.
Also known as hox genes.

47
Q

Define the term “homeodomain”.

A

A conserved motif of 60 amino acids found in all homeobox proteins. It is the part of the protein that binds to DNA, allowing the protein to act as a transcriptional regulator.

48
Q

Name the 3 kingdoms which all have very similar homeobox genes.

A

Plants, animals and fungi.

49
Q

Describe how homeobox genes are similar across life and how they differ, and define the term “highly conserved”.

A

Highly conserved = very similar.

50
Q

Define the term “hox genes”.

A

A group of homeobox genes that are only present in animals.

51
Q

Describe the role of Hox genes in controlling the body plan of animals.

A
  • Responsible for the correct development of body plans. Correct body plans ensures organism’s body parts are positioned correctly.
  • The order in which the hox genes are expressed along the chromosome is the order in which their effects are expressed in the organism.
  • Humans have 39 Hox genes.
52
Q

Describe the layout of living organisms.

A
  • Body plans are represented in cross-sections through the organism showing the fundamental arrangement of tissue layers.
  • Diplobastic animals have two primary layers of tissues and tripoblastic animals have three.
  • Common feature of animals is that they are segmented (i.e. rings on a ring worm or back bone of vertebrates.)
  • These segments have multiplied over time and have specialised to perform different functions.
  • Hox genes in the head control the development of wings, limbs or ribs.
  • Somites are segments in the embryo which lead to the development of individual vertebrae and assosciated structures.
  • Somites are directed by hox genes to develop in a particular way depending on their position in the sequence.
53
Q

What are the different types of symmetry in animal body shapes?

A
  • Radial symmetry is shown in dipoblastic animals (e.g. Jellyfish). They have no right or left sides, only a top and bottom.
  • Bilateral symmetry is seen in most animals. Organism have both left and right sides and a head and tail.
  • Asymmetry is seen in sponges which have no lines of symmetry.
54
Q

Define the term “apoptosis”.

A

Programmed and controlled cell death important in controlling the body form and in the removal of damaged or diseased cells.

55
Q

Outline the process of apoptosis.

A
  • Apoptosis shapes different body parts by removing unwanted cells and tissues.
  • Cells undergoing apoptosis can also release chemical signals which stimulate mitosis and cell proliferation leading to remodelling of tissues.
56
Q

Describe the process of apoptosis.

A

1) Enzymes inside the cell breakdown important cell components such as proteins in the cytoplasm and DNA in the nucleus.
2) As the cell’s contents are broken down it begins to skrink and breaks up into fragments.
3) The cell fragments are engulfed by phagocytes and digested.

57
Q

Explain how apoptosis is controlled.

A
  • Hox genes regulate apoptosis.
  • During development, genes than control apoptosis and genes that control mitosis are switched on and off in the appropriate cells.
  • This means some cells die whilst some new cells are produced and the correct body plan develops.
58
Q

Describe the role of mitosis and apoptosis in growth and development.

A

Mitosis and differentiation create the bulk of body parts and then apoptosis refines parts by removing the unwanted structures.
Apoptosis also removes damaged or diseased cells.

59
Q

Define the term “stress” in relation to homeostasis and describe the factors that may influence the rate of mitosis or apoptosis.

A
  • Stress = the condition produced when the homeostatic balance within an organism is upset.
  • Apoptosis/ mitosis respond to external and internal stimuli.
  • External stimuli (such as stress due to a lack a particular nutrient) could results in gene expression that prevents cells from undergoing mitosis.
  • External stimulus, such as attack of a pathogen, could lead to gene expression which triggers apoptosis.
  • Internal stimulus could be DNA damage. If DNA damaged is detected during the cell cycle, this can result in the expression of genes which cause the cycle to be paused and even trigger apoptosis.