6.1.1 Cellular control

Question 1

What are mutations?

Answer 1

-any change in the sequence of bases in the DNA

Question 2

How could a mutation lead to a non-functional protein?

Answer 2

-a change in the codon could mean that it codes for a different amino acid, which would lead to a change in the primary structure of the protein
-a change in the primary strucute may change the 3D(tertiary) of the protein –} different variable region= different interactions
-if the protein is an enzyme, it could lead to a change in the active site of the enzyme, meaning it can no longer bind to its specific substrate

Question 3

Why can some mutations not have an effect on the protein?

Answer 3

-the degenerate nature of the genetic code may mean that the new codon still codes for the same amino acid–} no change in the protein synthesised

Question 4

What are the 3 types of mutations?

Answer 4

-substitution= one or more bases are swapped for another base within the triplet code
-deletion= one or more bases are removed
-insertion= one or more bases are added

Question 5

Point mutations

Answer 5

-there are 3 types of substitution:
- silent mutation= a change in the base of the triplet, but the AA that the triplet codes for doesn’t change because genetic code is degenerate
- nonsense mutations= changes the triplet code to a stop codon which stops translation–} truncated protein
- missense mutation= change in the triplet code so that it codes for a different AA(can either have a similar function to the original AA or can be not involved with the protein’s function)

Question 6

Indel mutations

Answer 6

-mutations where there is an insertion/deletion of bases–} causes a frameshift
-this changes the way the rest of base sequences after that triplet code is read
-the earlier a frameshift mutation appears in a base sequence, the more AA’s affected + the greater the mutation’s effect on the protein

Question 7

Mutations that have no effect on the organism

Answer 7

-i.e eye colour
-no effect on the phenotype of an organism because normally functioning proteins are still synthesised

Question 8

Harmful mutations

Answer 8

-the phenotype of an organism is affected negatively because proteins are not synthesised/non functional proteins are synthesised
-i.e cystic fibrosis(caused by a deletion of 3 bases, leads to excess mucus production which affects lungs), sickle cell anaemia(flattens out RBCs and makes them spiky), down syndrome(chromosomal mutation)

Question 9

Beneficial mutations

Answer 9

-can result in a useful characteristic for an organism’s phenotype i.e a mutation in proteins on the cell surface membrane meaning HIV cannot bind and enter cells, skin colour(melanin absorbs sunlight)

Question 10

What is gene expression?

Answer 10

-the synthesis of the protein that the gene codes for

Question 11

Gene regulation

Answer 11

-every cell in an organism contains the same DNA regardless of its structure or function
-the expression of genes + rate of protein synthesis has to be regulated to prevent wastefulness–} done by a gene being switched on or off to increase/decrease the rate of protein synthesis

Question 12

Compare the regulation response in eukaryotes and prokaryotes

Answer 12

-multicellular organisms have to respond to stimuli from both external and internal stimuli whereas prokaryotes only respond to external stimuli to regulate gene expression

Question 13

What are the levels of gene regulation to control gene expression?

Answer 13

-transcriptional= genes can be turned on/off
-post-transcriptional= the editing of
primary mRNA and the removal of introns to
produce mature mRNA
-translational= translation can be stopped/started
-post-translational level= the activation of
proteins by cyclic AMP

Question 14

Transcriptional control in eukaryotes: transcription factors

Answer 14

-proteins that regulate gene expression by binding to the promoter region in the DNA (upstream to the gene) so that RNA polymerase can bind + initiate transcription/switch on the gene

Question 15

Transcription factor activation

Answer 15

-TF’s are usually in a deactivated state(their structure doesn’t allow them to bind to the promoter)
-must be activators in order to start transcription
(the shape of a TF determines whether it can bind or not, can be altered by the binding of some molecules i.e hormones/sugars)

Question 16

Transcriptional control: Chromatin remodelling

Answer 16

-DNA is wrapped around histones due to histones being positively charged
-chromatin can be modelled to make the genes more or less easily accessible to RNA polymerase by changing the charge difference between DNA and histones–} either more tightly or loosely bound

Question 17

Post-transcriptional control in eukaryotes: RNA editing

Answer 17

-primary mRNA(product of transcription) is modified to produce mature mRNA before it can bind to a ribosome and code for protein synthesis
-primary mRNA has both introns and exons—} introns are removed from pmRNA strands via splicing in the nucleus(hydrolysis of phosphodiester bonds)
-the mRNA can then leave the nucleus and travel to the cytoplasm
-genes can be regulated via the rate of mRNA editing= the faster it occurs, the more mature mRNA there is available for transcription
(nucleotide sequence may also substitution, insertion or deletion which may result in the synthesis of different function proteins—} increase the range of proteins that can be produced from one mRNA)

Question 18

Post-transcriptional control: Alternative splicing

Answer 18

-the same section of mRNA can be spliced together to code for different proteins

Question 19

Why is post-translational control needed?

Answer 19

-some proteins need to modified after they are synthesised–} done by a molecule binding to the protein to to activate it(cell signalling)

Question 20

Cell signalling in post-translational control

Answer 20

-first messenger with complimentary shape binds to specific receptor on cell surface membrane of target cell
-this triggers production of the second messenger,** cAMP** which leads activation of a protein in response to external stimuli via changing its tertiary structure

Question 21

example: activation of protein kinase A by cAMP

Answer 21

(PKA is an enzyme made of 4 subunits that can catalyse phosphorylation)
-when cAMP isn’t bound, thee 4 units are bound together and inactive
-when cAMP binds, it causes a change in the enzyme’s tertiary structure, releasing the active subunits–} PKA is now active

Question 22

How is prokaryotic regulation different to that of eukaryotes?

Answer 22

-prokaryotes do not have introns, therefore usually have smaller genes
-no nucleus means that transcription and translation happen at the same time
-there are no transcription factors in prokaryotes

Question 23

What is an operon?

Answer 23

-group of structural genes that are under the control of the same regulatory mechanism + are expressed at the same time
-i.e all genes needed to digest lactose are expressed together

Question 24

Why are operons useful?

Answer 24

-efficient way for unicellular organisms to save resources by switching genes on or off
-can respond quicker through transcription of multiple proteins at once as all the genes share 1 promoter region

Question 25

Describe the structure of an operon

Answer 25

consists of:
-a promoter(DNA sequence that RNA polymerase can bind to)
-a regulatory gene(codes for activator/response)
-an operator region
-structural genes(all transcribed together)

Question 26

Why does Lac Operon exist in E.coli?

Answer 26

-E is a bacterium that metabolises glucose and uses it as a respiratory substrate–} if glucose is in short supply, it can use lactose as a respiratory substrate

Question 27

Describe makeup of a Lac Operon

Answer 27

-group of 3 structural genes: lacZ, lacY and lacA–} transcribed onto a single long molecule of mRNA
-lacI (a regulatory gene) is located near the operon + codes for a repressor protein
-lacO is the operator region

Question 28

What do the structural genes code for?

Answer 28

• lacZ codes for B-galactosidase(catalyses hydrolysis of lactose into glucose + galactose)
• lacY codes for lactose permidase which makes cell more permeable to lactose
• lacA codes for transacetylase

Question 29

What happens when E.coli is grown on agar jelly containing lactose AND glucose, and how does this occur?

Answer 29

• gene expression of operon is switched on

-lac I codes for expression of repressor protein
-some lactose enters the cell and binds to second binding site of repressor protein–} causes a change in tertiary structure so it can no longer bind with the lacO and block the promoter region
-as a result, RNA polymerase can bind to the promoter region so that transcription of the 3 structural genes take place + enzymes are synthesised

Question 30

What happens when E.coli is grown on agar jelly containing ONLY glucose, and how does this occur?

Answer 30

• turn off gene expression of operon

-lac I codes for expression of repressor protein(has 2 binding sites)–} one site binds to the lacO, which prevents the enzyme RNA polymerase from binding to the promoter region due to the repressor protein being large in size and blocking promoter
-means that transcription will not occur and enzymes are not synthesised

Question 31

Role of cyclic AMP

Answer 31

-the binding of RNA polymerase only results in a relatively slow rate of transcription–} needs to be up-regulated to produce required quantity of enzymes to efficiently metabolise lactose
-achieved by the binding of CRP(cAMP repressor protein) when it is bound to cAMP
-transport of glucose into E.coli decreases levels of cAMP which reduces transcription(inversely proportional)

Question 32

What happens when E.coli is grown on agar jelly containing only lactose (and glucose is absent), and how does this occur?

Answer 32

-increase gene expression of operon

-lack of glucose means that more cAMP is present (inversely proportional so that transcription can still occur using lactose as respiratory substrate)
-cAMP can bind to CRP which alters the tertiary structure of CRP so that it can bind to the promoter region of the operon
-this increases the rate of transcription of the structural genes as it helps with the initial binding of RNA polymerase to the promoter

Question 33

What are body plans?

Answer 33

-the general structure of an organism
-proteins control the development of a body plan–} help set up the basic body plan so genes are expressed in the right place

Question 34

What are homeobox genes?

Answer 34

-a group of regulatory genes which all contain a homeobox(section of DNA which codes for part of a protein called the homeodomain)
-the protein coded for is highly conserved in plants, animals and fungi(DNA sequence is almost identical)–} mutations in homeobox genes would be embryo-lethal and are therefore remain unchanged during evolution by natural selection

Question 35

How are homeobox genes regulatory?

Answer 35

-the homeodomain of a protein binds to DNA at the start of developmental genes, which enables the protein to work as a TF–} to activate or repress transcription + so altering the production of proteins involved in the development of the body plan

Question 36

What are hox genes?

Answer 36

-homeobox genes only present in animals
-normally found in gene clusters
-responsible for the correct positioning of body parts–} the order in which genes appear on the chromosome is the order in which the gene is expressed in the organism i.e 1st hox in the head

Question 37

What is apoptosis?

Answer 37

-highly controlled process where some cells will die and break down as a part of development
-when this is triggered, the cell breaks down in a series of steps

Question 38

Describe the steps of apoptosis

Answer 38

1. the DNA in the nucleus + proteins are degraded(broken down) by enzymes
2. other cell components i.e mitochondria + cytoskeleton are also broken down
3. as the cell contents are broken down, it begins to shrink and break up into fragments
4. ‘blebs’ develop on cell surface and are engulfed by phagocytes + are digested

Question 39

What is the role of apoptosis and mitosis in development?

Answer 39

-mitosis + differentiation create the bulk of the body parts and apoptosis refines the parts by removing unwanted structures
-cells undergoing apoptosis can also release chemical signals which stimulate mitosis + cell proliferation leading to the remodelling of tissues–} both take cues from internal + external environment

Question 40

Examples of internal + external stimuli that trigger cell cycle/apoptosis

Answer 40

-internal= DNA damage–} can result in the expression of genes if detected during cell cycle, which causes cycle to pause and can trigger apoptosis
-external= stress caused by a lack of nutrient availability–} could result in gene expression that prevents cells from undergoing mitosis
-attack by a pathogen can trigger apoptosis