U3 AoS1 topics I had difficulty remebering Flashcards
what are the advantages of degeneracy? (genetic code)
If a mutation is present within an organism’s DNA, chances are that a single nucleotide difference may not affect the resulting amino acid produced by the DNA triplet.
identify and sum up the three steps of transcription
initiation- RNA polymerase binds to promoter region of DNA and unzips the DNA.
elongation- RNA polymerase reads 3’ to 5’ template strand & makes complementary strand of pre-mRNA. Nucelotides are added to growing 3’ end.
termination- After RNA polymerase moves downstream past the coding region,it reaches the termination sequence and detaches, releasing pre-mRNA strand and DNA zips back up.
identify and briefly explain the steps of RNA processing
capping- 5’ methyl cap with altered G base is added to increase stability and help ribosome to bind to RNA during translation.
poly-A-tail- a tail consisting of many A bases is added to the 3’ end of mRNA strand to facilitate export and increase stability.
splicing- spliceosomes remove introns. Recognises introns GU at 5’ and AG at 3’.
- alternative splicing- additonal exons are removed and one gene can encode for many proteins.
identify and explain the steps of translation
initiation- small ribosome subunit and first tRNA molecule dock onto methyl cap which from there scan for start codon to bind onto. Once found, Large subunit binds
elongation- first tRNA arrives in P site, next complementary tRNA fills A site and anticodons of tRNA bind to codon of mRNA and amino acids join between peptide bonds. Empty tRNA leaves P site and ribosome moves in 3’ direction.
termination- STOP codon is reached and release factor enters A site, signalling for all components to disassociate.
7 functions of proteins
HORMONE- eg insulin; regulate body activity
RECEPTORS- eg insulin receptor; respond to stimuli
TRANSPORT- eg haemoglobin; carry other molecules
IMMUNOGLOBIN eg antibodies; defence against disease
STRUCTURAL- eg keratin; fibrous support tissue
CONTRACTILE- eg myosin; facilitate muscle movement
ENZYMES- eg restriction endonuclease; catalyse chemical reactions
summarise three main forms of RNA
tRNA- carries amino acid to ribosome and anticodon binds to codon of mRNA during translation to play key role in protein synthesis. Release amino acids to form polypeptide chain
mRNA- produced during transcription & RNA processing in the nucleus and carries codons to ribosome to send a message for the production of specific polypeptide chains.
rRNA- with proteins, makes up ribosomes.
describe the protein secretory pathway
RER- in the RER, proteins are produced through ribosomes and processing occurs (for example addition of sugar groups and assembling complex proteins)
Golgi apparatus- Protein modification. Processes and packages proteins into secretory vesicles for export.
vesicles- carry proteins out of cell during exocytosis by fusing to membrane and discharging contents out of the cell.
outline the steps in PCR
Denaturing- DNA strands are separated, raised temperature (94 degrees for one minute) cuases H bonds to break
Annealing-short segments of DNA (primers) are added. Primers bind to 3’ end and initiate polymerisation (55 degrees for tow minutes)
Extending- taq polymerase uses primers as a starting point and extends the primers, synthesising DNA by adding nucleotides. 72 degrees for one minute.
outline CRISPR Cas 9
exposure to virus- recognition of vDNA sequence (PAM sequences) section of vDNA (protospacer) is cut out by cas enzymes
incorporation protospacer becomes a spacer in CRISPR
creation of cas 9 complex cas 9 is transcribed and translated, and CRISPR spacer is transcribed to get gRNA
extermination cas 9 scans DNA for PAM sequence, when found it docks onto it. Uses gRNA to check if DNA through complimentary bases (if yes, both strands of DNA are cut by enzyme)
if ligases repair but with occasional error the viral gene is non functional
outline gene therapy
1- sgRNA is created that matches target and is added to a mix with cas 9
2 cell is injected with mix
3 PAM sequence identified and cut out
4 faulty gene can be replaced by corrected DNA or new gene
5 cells attempt to repair silences targeted gene
outline recombination
gene of interest is vut and so is the plasmid with SAME restriction enzyme
join gene and plasmid with DNA ligase
check if reporter gene is disrupted to see if gene has been taken up.
outline transformation
bacteria either goes through heat shock or electroporation to fit large plasmid inside membrane
check if bacteria is resistant to antibiotic. If it has taken up the plasmid, it will be resistant.
Using your knowledge of CRISPR-Cas9, explain how this may be examined as a possible mechanism to treat Huntington’s disease.
2 marks
Gene therapy could be used as it targets particular PAM sequences and cuts them. It could cut out the mutated gene and the cell’s attempt to repair it would effectively silence the gene.
this answer is more generalised. specific answer to huntingtons below
CRISPR-Cas9 can specifically target the HD gene. Cas9 can cut on either side of the excessive CAG repeats, allowing these extra repeats to be removed (1 mark). The DNA can then repair without the extra repeats (1 mark).
Why is PCR important in the process of diagnosing Huntington’s disease?
PCR is important in the disagnosis as it amplifies the DNA so scientists can recognise the presence or absence of a mutation in the gene during gel electrophoresis
Outline how Huntington’s disease can affect the primary, secondary and tertiary structure of the huntingtin protein.
primary- the linear sequence has been changed to different amino acids and is longer.
secondary- the folds (alpha helices beta pleated sheets and random coils) may not be able to form due to changes in R groups.
tertiary- 3D folding will be effected due to presence of extra amino acids, leading to different bonds forming and potentially causing loss of function.
SEE JACPLUS FOR PICTURE
Which child or children will likely develop Huntington’s disease? How do you know?
The children could inherit an allele of 48 repeats or the 18 repeats allele from their mother and the 17 or 19 repeats alleles from their father. II-1 and II-3 are likely to be affected as they both have a longer band. This is because they have a longer sequence of nucleotides in their DNA when affected, as there is a larger number of repeats, and so the when the restriction enzyme cuts the DNA, one fragment will be much longer than the other. Because of the presence of the longer bands, forty or more copies of the repeat, it is likely they will be affected by Hungtinton’s disease.
3 marks- explain possibilites -those with disease & -why this is answer
SEE JACPLUS
When the second child has their own child, their child (at age 16) chooses to have testing for Huntington’s disease. What implications may these results have for their parent?
As Huntington’s disease is dominant, if the child tests positive for the H allele this must be inherited from their parent (likely the one with the family history) (1 mark). Therefore, the parent may find out by association that they also have the gene and will develop Huntington’s, despite choosing to not find out (1 mark).
Where is tRNA found in a cell?
cytosol
Describe the role of tRNA.
tRNA brings specific amino acids to ribosomes, where they are assembled into polypeptide chains. The tRNA anticodon pairs with the mRNA codon during translation at the ribosome. In this way, amino acids are assembled in an order determined by the mRNA.
Describe a distinctive property of a fibrous protein and explain how this property is due to the arrangement of its polypeptides.
Strength: this is determined by the parallel arrangement of the polypeptides into long fibres and sheets.
Explain what is meant by gene regulation.
Gene regulation means only expressing genes when the cell requires the product.
outline the function of mRNA
mRNA: determines the order amino acids are assembled at the ribosome or carries the information that determines the order amino acids are assembled, from the nucleus of eukaryotic cells to the ribosome.
A patient with tuberculosis had a persistent cough, fever and weakness. The patient was treated with antibiotic X and began to improve after finishing the course of antibiotics. However the cough and other symptoms returned.
a
Explain the most likely reason why the course of antibiotics was not successful.
The course of antibiotics was not successful because some of the original population were resistant to the antibiotic. When the patient was first treated with antibiotic most bacteria died and the symptoms began to disappear. The few remaining resistant bacteria reproduced and their offspring inherited resistance to the antibiotic but it took time for their numbers to increase to the level that symptoms began to appear again.
Transgenic bacteria have been produced by inserting a gene or genes from another organism.
Particular species of transgenic bacteria have been developed to break down plastic compounds. Before releasing these bacteria into the environment scientists made extensive studies regarding the conditions in which the bacteria grew.
b
Describe one important reason for this extensive study.
Transgenic bacteria usually contain antibiotic resistant alleles. Scientists must make sure that these bacteria are not harmful as using an antibiotic to control them will be ineffective.
